mirror of
https://github.com/youngyangyang04/leetcode-master.git
synced 2025-07-08 16:54:50 +08:00
Merge branch 'youngyangyang04:master' into master
This commit is contained in:
MrYoungg
49
README.md
49
README.md
@ -372,25 +372,38 @@
|
||||
|
||||
通知：开始更新图论内容，图论部分还没有其他语言版本，欢迎录友们提交PR，成为contributor
|
||||
|
||||
### 深搜广搜
|
||||
1. [图论：理论基础](./problems/kamacoder/图论理论基础.md)
|
||||
2. [图论：深度优先搜索理论基础](./problems/kamacoder/图论深搜理论基础.md)
|
||||
3. [图论：所有可达路径](./problems/kamacoder/0098.所有可达路径.md)
|
||||
4. [图论：广度优先搜索理论基础](./problems/kamacoder/图论广搜理论基础.md)
|
||||
5. [图论：岛屿数量.深搜版](./problems/kamacoder/0099.岛屿的数量深搜.md)
|
||||
6. [图论：岛屿数量.广搜版](./problems/kamacoder/0099.岛屿的数量广搜.md)
|
||||
7. [图论：岛屿的最大面积](./problems/kamacoder/0100.岛屿的最大面积.md)
|
||||
8. [图论：孤岛的总面积](./problems/kamacoder/0101.孤岛的总面积.md)
|
||||
9. [图论：沉没孤岛](./problems/kamacoder/0102.沉没孤岛.md)
|
||||
10. [图论：水流问题](./problems/kamacoder/0103.水流问题.md)
|
||||
11. [图论：建造最大岛屿](./problems/kamacoder/0104.建造最大岛屿.md)
|
||||
12. [图论：字符串接龙](./problems/kamacoder/0110.字符串接龙.md)
|
||||
13. [图论：有向图的完全可达性](./problems/kamacoder/0105.有向图的完全可达性.md)
|
||||
14. [图论：岛屿的周长](./problems/kamacoder/0106.岛屿的周长.md)
|
||||
15. [图论：并查集理论基础](./problems/kamacoder/图论并查集理论基础.md)
|
||||
16. [图论：寻找存在的路径](./problems/kamacoder/0107.寻找存在的路径.md)
|
||||
17. [图论：冗余连接](./problems/kamacoder/0108.冗余连接.md)
|
||||
18. [图论：冗余连接II](./problems/kamacoder/0109.冗余连接II.md)
|
||||
19. [图论：最小生成树之prim](./problems/kamacoder/0053.寻宝-prim.md)
|
||||
20. [图论：最小生成树之kruskal](./problems/kamacoder/0053.寻宝-Kruskal.md)
|
||||
21. [图论：拓扑排序](./problems/kamacoder/0117.软件构建.md)
|
||||
22. [图论：dijkstra（朴素版）](./problems/kamacoder/0047.参会dijkstra朴素.md)
|
||||
23. [图论：dijkstra（堆优化版）](./problems/kamacoder/0047.参会dijkstra堆.md)
|
||||
24. [图论：Bellman_ford 算法](./problems/kamacoder/0094.城市间货物运输I.md)
|
||||
25. [图论：Bellman_ford 队列优化算法（又名SPFA）](./problems/kamacoder/0094.城市间货物运输I-SPFA.md)
|
||||
26. [图论：Bellman_ford之判断负权回路](./problems/kamacoder/0095.城市间货物运输II.md)
|
||||
27. [图论：Bellman_ford之单源有限最短路](./problems/kamacoder/0095.城市间货物运输II.md)
|
||||
28. [图论：Floyd 算法](./problems/kamacoder/0097.小明逛公园.md)
|
||||
29. [图论：A * 算法](./problems/kamacoder/0126.骑士的攻击astar.md)
|
||||
30. [图论：最短路算法总结篇](./problems/kamacoder/最短路问题总结篇.md)
|
||||
31. [图论：图论总结篇](./problems/kamacoder/图论总结篇.md)
|
||||
|
||||
* [图论：深度优先搜索理论基础](./problems/图论深搜理论基础.md)
|
||||
* [图论：797.所有可能的路径](./problems/0797.所有可能的路径.md)
|
||||
* [图论：广度优先搜索理论基础](./problems/图论广搜理论基础.md)
|
||||
* [图论：200.岛屿数量.深搜版](./problems/0200.岛屿数量.深搜版.md)
|
||||
* [图论：200.岛屿数量.广搜版](./problems/0200.岛屿数量.广搜版.md)
|
||||
* [图论：695.岛屿的最大面积](./problems/0695.岛屿的最大面积.md)
|
||||
* [图论：1020.飞地的数量](./problems/1020.飞地的数量.md)
|
||||
* [图论：130.被围绕的区域](./problems/0130.被围绕的区域.md)
|
||||
* [图论：417.太平洋大西洋水流问题](./problems/0417.太平洋大西洋水流问题.md)
|
||||
* [图论：827.最大人工岛](./problems/0827.最大人工岛.md)
|
||||
* [图论：127. 单词接龙](./problems/0127.单词接龙.md)
|
||||
* [图论：841.钥匙和房间](./problems/0841.钥匙和房间.md)
|
||||
* [图论：463. 岛屿的周长](./problems/0463.岛屿的周长.md)
|
||||
* [图论：并查集理论基础](./problems/图论并查集理论基础.md)
|
||||
* [图论：1971. 寻找图中是否存在路径](./problems/1971.寻找图中是否存在路径.md)
|
||||
* [图论：684.冗余连接](./problems/0684.冗余连接.md)
|
||||
* [图论：685.冗余连接II](./problems/0685.冗余连接II.md)
|
||||
|
||||
（持续更新中....）
|
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|
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|
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@ -178,13 +178,13 @@ public:
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// 记录从大西洋出发，可以遍历的节点
|
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vector<vector<bool>> atlantic = vector<vector<bool>>(n, vector<bool>(m, false));
|
||||
|
||||
// 从最上最下行的节点出发，向高处遍历
|
||||
// 从最左最右列的节点出发，向高处遍历
|
||||
for (int i = 0; i < n; i++) {
|
||||
dfs (heights, pacific, i, 0); // 遍历最左列，接触太平洋
|
||||
dfs (heights, atlantic, i, m - 1); // 遍历最右列，接触大西
|
||||
}
|
||||
|
||||
// 从最左最右列的节点出发，向高处遍历
|
||||
// 从最上最下行的节点出发，向高处遍历
|
||||
for (int j = 0; j < m; j++) {
|
||||
dfs (heights, pacific, 0, j); // 遍历最上行，接触太平洋
|
||||
dfs (heights, atlantic, n - 1, j); // 遍历最下行，接触大西洋
|
||||
@ -297,6 +297,73 @@ class Solution {
|
||||
}
|
||||
```
|
||||
|
||||
```Java
|
||||
class Solution {
|
||||
|
||||
// 和Carl题解更加符合的Java DFS
|
||||
private int[][] directions = {{-1, 0}, {1, 0}, {0, 1}, {0, -1}};
|
||||
|
||||
/**
|
||||
* @param heights 题目给定的二维数组
|
||||
* @param m 当前位置的行号
|
||||
* @param n 当前位置的列号
|
||||
* @param visited 记录这个位置可以到哪条河
|
||||
*/
|
||||
|
||||
public void dfs(int[][] heights, boolean[][] visited, int m, int n){
|
||||
if(visited[m][n]) return;
|
||||
visited[m][n] = true;
|
||||
|
||||
for(int[] dir: directions){
|
||||
int nextm = m + dir[0];
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int nextn = n + dir[1];
|
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//出了2D array的边界，continue
|
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if(nextm < 0||nextm == heights.length||nextn <0||nextn== heights[0].length) continue;
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//下一个位置比当下位置还要低，跳过，继续找下一个更高的位置
|
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if(heights[m][n] > heights[nextm][nextn]) continue;
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dfs(heights, visited, nextm, nextn);
|
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}
|
||||
}
|
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|
||||
|
||||
public List<List<Integer>> pacificAtlantic(int[][] heights) {
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int m = heights.length;
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int n = heights[0].length;
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|
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// 记录从太平洋边出发，可以遍历的节点
|
||||
boolean[][] pacific = new boolean[m][n];
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// 记录从大西洋出发，可以遍历的节点
|
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boolean[][] atlantic = new boolean[m][n];
|
||||
|
||||
// 从最左最右列的节点出发，向高处遍历
|
||||
for(int i = 0; i<m; i++){
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dfs(heights, pacific, i, 0); //遍历pacific最左边
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dfs(heights, atlantic, i, n-1); //遍历atlantic最右边
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}
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|
||||
// 从最上最下行的节点出发，向高处遍历
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for(int j = 0; j<n; j++){
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dfs(heights, pacific, 0, j); //遍历pacific最上边
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dfs(heights, atlantic, m-1, j); //遍历atlantic最下边
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}
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|
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List<List<Integer>> result = new ArrayList<>();
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for(int a = 0; a<m; a++){
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for(int b = 0; b<n; b++){
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// 如果这个节点，从太平洋和大西洋出发都遍历过，就是结果
|
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if(pacific[a][b] && atlantic[a][b]){
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List<Integer> pair = new ArrayList<>();
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pair.add(a);
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pair.add(b);
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result.add(pair);
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}
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||||
}
|
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}
|
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return result;
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}
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}
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```
|
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|
||||
广度优先遍历：
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||||
|
||||
```Java
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@ -159,7 +159,89 @@ public:
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* 时间复杂度: O(kmn)，k 为strs的长度
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* 空间复杂度: O(mn)
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||||
|
||||
C++:
|
||||
使用三维数组的版本
|
||||
|
||||
```CPP
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||||
class Solution {
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||||
public:
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||||
int findMaxForm(vector<string>& strs, int m, int n) {
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int num_of_str = strs.size();
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vector<vector<vector<int>>> dp(num_of_str, vector<vector<int>>(m + 1,vector<int>(n + 1, 0)));
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/* dp[i][j][k] represents, if choosing items among strs[0] to strs[i] to form a subset,
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what is the maximum size of this subset such that there are no more than m 0's and n 1's in this subset.
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Each entry of dp[i][j][k] is initialized with 0
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|
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transition formula:
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using x[i] to indicates the number of 0's in strs[i]
|
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using y[i] to indicates the number of 1's in strs[i]
|
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|
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dp[i][j][k] = max(dp[i-1][j][k], dp[i-1][j - x[i]][k - y[i]] + 1)
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|
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*/
|
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|
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|
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// num_of_zeros records the number of 0's for each str
|
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// num_of_ones records the number of 1's for each str
|
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// find the number of 0's and the number of 1's for each str in strs
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vector<int> num_of_zeros;
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vector<int> num_of_ones;
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for (auto& str : strs){
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int count_of_zero = 0;
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int count_of_one = 0;
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for (char &c : str){
|
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if(c == '0') count_of_zero ++;
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else count_of_one ++;
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}
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num_of_zeros.push_back(count_of_zero);
|
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num_of_ones.push_back(count_of_one);
|
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|
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}
|
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|
||||
|
||||
// num_of_zeros[0] indicates the number of 0's for str[0]
|
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// num_of_ones[0] indiates the number of 1's for str[1]
|
||||
|
||||
// initialize the 1st plane of dp[i][j][k], i.e., dp[0][j][k]
|
||||
// if num_of_zeros[0] > m or num_of_ones[0] > n, no need to further initialize dp[0][j][k],
|
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// because they have been intialized to 0 previously
|
||||
if(num_of_zeros[0] <= m && num_of_ones[0] <= n){
|
||||
// for j < num_of_zeros[0] or k < num_of_ones[0], dp[0][j][k] = 0
|
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for(int j = num_of_zeros[0]; j <= m; j++){
|
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for(int k = num_of_ones[0]; k <= n; k++){
|
||||
dp[0][j][k] = 1;
|
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}
|
||||
}
|
||||
}
|
||||
|
||||
/* if j - num_of_zeros[i] >= 0 and k - num_of_ones[i] >= 0:
|
||||
dp[i][j][k] = max(dp[i-1][j][k], dp[i-1][j - num_of_zeros[i]][k - num_of_ones[i]] + 1)
|
||||
else:
|
||||
dp[i][j][k] = dp[i-1][j][k]
|
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*/
|
||||
|
||||
for (int i = 1; i < num_of_str; i++){
|
||||
int count_of_zeros = num_of_zeros[i];
|
||||
int count_of_ones = num_of_ones[i];
|
||||
for (int j = 0; j <= m; j++){
|
||||
for (int k = 0; k <= n; k++){
|
||||
if( j < count_of_zeros || k < count_of_ones){
|
||||
dp[i][j][k] = dp[i-1][j][k];
|
||||
}else{
|
||||
dp[i][j][k] = max(dp[i-1][j][k], dp[i-1][j - count_of_zeros][k - count_of_ones] + 1);
|
||||
}
|
||||
}
|
||||
}
|
||||
|
||||
}
|
||||
|
||||
return dp[num_of_str-1][m][n];
|
||||
|
||||
}
|
||||
};
|
||||
```
|
||||
|
||||
## 总结
|
||||
|
||||
|
||||
@ -1,4 +1,6 @@
|
||||
|
||||
<p align="center"><strong><a href="./qita/join.md">参与本项目</a>，贡献其他语言版本的代码，拥抱开源，让更多学习算法的小伙伴们受益！</strong></p>
|
||||
|
||||
# dijkstra（堆优化版）精讲
|
||||
|
||||
[卡码网：47. 参加科学大会](https://kamacoder.com/problempage.php?pid=1047)
|
||||
|
||||
@ -1,4 +1,6 @@
|
||||
|
||||
<p align="center"><strong><a href="./qita/join.md">参与本项目</a>，贡献其他语言版本的代码，拥抱开源，让更多学习算法的小伙伴们受益！</strong></p>
|
||||
|
||||
# dijkstra（朴素版）精讲
|
||||
|
||||
[卡码网：47. 参加科学大会](https://kamacoder.com/problempage.php?pid=1047)
|
||||
|
||||
@ -1,4 +1,6 @@
|
||||
|
||||
<p align="center"><strong><a href="./qita/join.md">参与本项目</a>，贡献其他语言版本的代码，拥抱开源，让更多学习算法的小伙伴们受益！</strong></p>
|
||||
|
||||
# kruskal算法精讲
|
||||
|
||||
[卡码网：53. 寻宝](https://kamacoder.com/problempage.php?pid=1053)
|
||||
|
||||
@ -1,4 +1,6 @@
|
||||
|
||||
<p align="center"><strong><a href="./qita/join.md">参与本项目</a>，贡献其他语言版本的代码，拥抱开源，让更多学习算法的小伙伴们受益！</strong></p>
|
||||
|
||||
# prim算法精讲
|
||||
|
||||
[卡码网：53. 寻宝](https://kamacoder.com/problempage.php?pid=1053)
|
||||
|
||||
@ -1,11 +1,4 @@
|
||||
|
||||
<p align="center">
|
||||
<a href="https://programmercarl.com/other/xunlianying.html" target="_blank">
|
||||
<img src="../pics/训练营.png" width="1000"/>
|
||||
</a>
|
||||
<p align="center"><strong><a href="https://mp.weixin.qq.com/s/tqCxrMEU-ajQumL1i8im9A">参与本项目</a>，贡献其他语言版本的代码，拥抱开源，让更多学习算法的小伙伴们收益！</strong></p>
|
||||
|
||||
|
||||
# 替换数字
|
||||
|
||||
[卡码网题目链接](https://kamacoder.com/problempage.php?pid=1064)
|
||||
@ -376,9 +369,3 @@ main();
|
||||
### Rust:
|
||||
|
||||
|
||||
|
||||
<p align="center">
|
||||
<a href="https://programmercarl.com/other/kstar.html" target="_blank">
|
||||
<img src="../pics/网站星球宣传海报.jpg" width="1000"/>
|
||||
</a>
|
||||
|
||||
|
||||
@ -1,9 +1,4 @@
|
||||
|
||||
<p align="center">
|
||||
<a href="https://programmercarl.com/other/xunlianying.html" target="_blank">
|
||||
<img src="../pics/训练营.png" width="1000"/>
|
||||
</a>
|
||||
<p align="center"><strong><a href="https://mp.weixin.qq.com/s/tqCxrMEU-ajQumL1i8im9A">参与本项目</a>，贡献其他语言版本的代码，拥抱开源，让更多学习算法的小伙伴们收益！</strong></p>
|
||||
|
||||
|
||||
# 右旋字符串
|
||||
@ -350,9 +345,3 @@ var reverseLeftWords = function(s, n) {
|
||||
### Rust:
|
||||
|
||||
|
||||
|
||||
|
||||
<p align="center">
|
||||
<a href="https://programmercarl.com/other/kstar.html" target="_blank">
|
||||
<img src="../pics/网站星球宣传海报.jpg" width="1000"/>
|
||||
</a>
|
||||
|
||||
@ -1,4 +1,6 @@
|
||||
|
||||
<p align="center"><strong><a href="./qita/join.md">参与本项目</a>，贡献其他语言版本的代码，拥抱开源，让更多学习算法的小伙伴们受益！</strong></p>
|
||||
|
||||
# Bellman_ford 队列优化算法（又名SPFA）
|
||||
|
||||
[卡码网：94. 城市间货物运输 I](https://kamacoder.com/problempage.php?pid=1152)
|
||||
|
||||
@ -1,4 +1,6 @@
|
||||
|
||||
<p align="center"><strong><a href="./qita/join.md">参与本项目</a>，贡献其他语言版本的代码，拥抱开源，让更多学习算法的小伙伴们受益！</strong></p>
|
||||
|
||||
# Bellman_ford 算法精讲
|
||||
|
||||
[卡码网：94. 城市间货物运输 I](https://kamacoder.com/problempage.php?pid=1152)
|
||||
|
||||
@ -1,4 +1,6 @@
|
||||
|
||||
<p align="center"><strong><a href="./qita/join.md">参与本项目</a>，贡献其他语言版本的代码，拥抱开源，让更多学习算法的小伙伴们受益！</strong></p>
|
||||
|
||||
# bellman_ford之判断负权回路
|
||||
|
||||
[卡码网：95. 城市间货物运输 II](https://kamacoder.com/problempage.php?pid=1153)
|
||||
|
||||
@ -1,4 +1,6 @@
|
||||
|
||||
<p align="center"><strong><a href="./qita/join.md">参与本项目</a>，贡献其他语言版本的代码，拥抱开源，让更多学习算法的小伙伴们受益！</strong></p>
|
||||
|
||||
# bellman_ford之单源有限最短路
|
||||
|
||||
[卡码网：96. 城市间货物运输 III](https://kamacoder.com/problempage.php?pid=1154)
|
||||
|
||||
@ -1,4 +1,6 @@
|
||||
|
||||
<p align="center"><strong><a href="./qita/join.md">参与本项目</a>，贡献其他语言版本的代码，拥抱开源，让更多学习算法的小伙伴们受益！</strong></p>
|
||||
|
||||
# Floyd 算法精讲
|
||||
|
||||
[卡码网：97. 小明逛公园](https://kamacoder.com/problempage.php?pid=1155)
|
||||
|
||||
@ -1,4 +1,6 @@
|
||||
|
||||
<p align="center"><strong><a href="./qita/join.md">参与本项目</a>，贡献其他语言版本的代码，拥抱开源，让更多学习算法的小伙伴们受益！</strong></p>
|
||||
|
||||
# 98. 所有可达路径
|
||||
|
||||
[卡码网题目链接（ACM模式）](https://kamacoder.com/problempage.php?pid=1170)
|
||||
|
||||
@ -1,4 +1,6 @@
|
||||
|
||||
<p align="center"><strong><a href="./qita/join.md">参与本项目</a>，贡献其他语言版本的代码，拥抱开源，让更多学习算法的小伙伴们受益！</strong></p>
|
||||
|
||||
# 99. 岛屿数量
|
||||
|
||||
[卡码网题目链接（ACM模式）](https://kamacoder.com/problempage.php?pid=1171)
|
||||
|
||||
@ -1,4 +1,6 @@
|
||||
|
||||
<p align="center"><strong><a href="./qita/join.md">参与本项目</a>，贡献其他语言版本的代码，拥抱开源，让更多学习算法的小伙伴们受益！</strong></p>
|
||||
|
||||
# 99. 岛屿数量
|
||||
|
||||
[卡码网题目链接（ACM模式）](https://kamacoder.com/problempage.php?pid=1171)
|
||||
|
||||
@ -1,4 +1,6 @@
|
||||
|
||||
<p align="center"><strong><a href="./qita/join.md">参与本项目</a>，贡献其他语言版本的代码，拥抱开源，让更多学习算法的小伙伴们受益！</strong></p>
|
||||
|
||||
# 100. 岛屿的最大面积
|
||||
|
||||
[卡码网题目链接（ACM模式）](https://kamacoder.com/problempage.php?pid=1172)
|
||||
|
||||
@ -1,4 +1,6 @@
|
||||
|
||||
<p align="center"><strong><a href="./qita/join.md">参与本项目</a>，贡献其他语言版本的代码，拥抱开源，让更多学习算法的小伙伴们受益！</strong></p>
|
||||
|
||||
# 101. 孤岛的总面积
|
||||
|
||||
[卡码网：101. 孤岛的总面积](https://kamacoder.com/problempage.php?pid=1173)
|
||||
|
||||
@ -1,4 +1,6 @@
|
||||
|
||||
<p align="center"><strong><a href="./qita/join.md">参与本项目</a>，贡献其他语言版本的代码，拥抱开源，让更多学习算法的小伙伴们受益！</strong></p>
|
||||
|
||||
# 102. 沉没孤岛
|
||||
|
||||
[卡码网题目链接（ACM模式）](https://kamacoder.com/problempage.php?pid=1174)
|
||||
|
||||
@ -1,4 +1,6 @@
|
||||
|
||||
<p align="center"><strong><a href="./qita/join.md">参与本项目</a>，贡献其他语言版本的代码，拥抱开源，让更多学习算法的小伙伴们受益！</strong></p>
|
||||
|
||||
# 103. 水流问题
|
||||
|
||||
[卡码网题目链接（ACM模式）](https://kamacoder.com/problempage.php?pid=1175)
|
||||
|
||||
@ -1,4 +1,6 @@
|
||||
|
||||
<p align="center"><strong><a href="./qita/join.md">参与本项目</a>，贡献其他语言版本的代码，拥抱开源，让更多学习算法的小伙伴们受益！</strong></p>
|
||||
|
||||
# 104.建造最大岛屿
|
||||
|
||||
[卡码网题目链接（ACM模式）](https://kamacoder.com/problempage.php?pid=1176)
|
||||
|
||||
@ -1,4 +1,6 @@
|
||||
|
||||
<p align="center"><strong><a href="./qita/join.md">参与本项目</a>，贡献其他语言版本的代码，拥抱开源，让更多学习算法的小伙伴们受益！</strong></p>
|
||||
|
||||
# 105.有向图的完全可达性
|
||||
|
||||
[卡码网题目链接（ACM模式）](https://kamacoder.com/problempage.php?pid=1177)
|
||||
@ -192,7 +194,7 @@ int main() {
|
||||
|
||||
**第二种写法注意有注释的地方是和写法一的区别**
|
||||
|
||||
```c++
|
||||
```CPP
|
||||
写法二：dfs处理下一个要访问的节点
|
||||
#include <iostream>
|
||||
#include <vector>
|
||||
|
||||
@ -1,4 +1,6 @@
|
||||
|
||||
<p align="center"><strong><a href="./qita/join.md">参与本项目</a>，贡献其他语言版本的代码，拥抱开源，让更多学习算法的小伙伴们受益！</strong></p>
|
||||
|
||||
# 106. 岛屿的周长
|
||||
|
||||
[卡码网题目链接（ACM模式）](https://kamacoder.com/problempage.php?pid=1178)
|
||||
|
||||
@ -1,4 +1,6 @@
|
||||
|
||||
<p align="center"><strong><a href="./qita/join.md">参与本项目</a>，贡献其他语言版本的代码，拥抱开源，让更多学习算法的小伙伴们受益！</strong></p>
|
||||
|
||||
# 107. 寻找存在的路径
|
||||
|
||||
[卡码网题目链接（ACM模式）](https://kamacoder.com/problempage.php?pid=1179)
|
||||
|
||||
@ -1,4 +1,6 @@
|
||||
|
||||
<p align="center"><strong><a href="./qita/join.md">参与本项目</a>，贡献其他语言版本的代码，拥抱开源，让更多学习算法的小伙伴们受益！</strong></p>
|
||||
|
||||
# 108. 冗余连接
|
||||
|
||||
[卡码网题目链接（ACM模式）](https://kamacoder.com/problempage.php?pid=1181)
|
||||
|
||||
@ -1,4 +1,6 @@
|
||||
|
||||
<p align="center"><strong><a href="./qita/join.md">参与本项目</a>，贡献其他语言版本的代码，拥抱开源，让更多学习算法的小伙伴们受益！</strong></p>
|
||||
|
||||
# 109. 冗余连接II
|
||||
|
||||
[卡码网题目链接（ACM模式）](https://kamacoder.com/problempage.php?pid=1182)
|
||||
|
||||
@ -1,4 +1,6 @@
|
||||
|
||||
<p align="center"><strong><a href="./qita/join.md">参与本项目</a>，贡献其他语言版本的代码，拥抱开源，让更多学习算法的小伙伴们受益！</strong></p>
|
||||
|
||||
# 110. 字符串接龙
|
||||
|
||||
[卡码网题目链接（ACM模式）](https://kamacoder.com/problempage.php?pid=1182)
|
||||
|
||||
@ -1,4 +1,6 @@
|
||||
|
||||
<p align="center"><strong><a href="./qita/join.md">参与本项目</a>，贡献其他语言版本的代码，拥抱开源，让更多学习算法的小伙伴们受益！</strong></p>
|
||||
|
||||
# 拓扑排序精讲
|
||||
|
||||
[卡码网：117. 软件构建](https://kamacoder.com/problempage.php?pid=1191)
|
||||
|
||||
@ -1,4 +1,6 @@
|
||||
|
||||
<p align="center"><strong><a href="./qita/join.md">参与本项目</a>，贡献其他语言版本的代码，拥抱开源，让更多学习算法的小伙伴们受益！</strong></p>
|
||||
|
||||
# 大数减法
|
||||
|
||||
本题测试数据超过int 和 longlong了，所以考察的使用 string 来模拟 两个大数的 加减操作。
|
||||
|
||||
49
problems/kamacoder/0129.小美的蛋糕切割.md
Normal file
49
problems/kamacoder/0129.小美的蛋糕切割.md
Normal file
@ -0,0 +1,49 @@
|
||||
|
||||
前缀和
|
||||
|
||||
```CPP
|
||||
|
||||
#include <iostream>
|
||||
#include <vector>
|
||||
#include <climits>
|
||||
|
||||
using namespace std;
|
||||
int main () {
|
||||
int n, m;
|
||||
cin >> n >> m;
|
||||
int sum = 0;
|
||||
vector<vector<int>> vec(n, vector<int>(m, 0)) ;
|
||||
for (int i = 0; i < n; i++) {
|
||||
for (int j = 0; j < m; j++) {
|
||||
cin >> vec[i][j];
|
||||
sum += vec[i][j];
|
||||
}
|
||||
}
|
||||
// 统计横向
|
||||
vector<int> horizontal(n, 0);
|
||||
for (int i = 0; i < n; i++) {
|
||||
for (int j = 0 ; j < m; j++) {
|
||||
horizontal[i] += vec[i][j];
|
||||
}
|
||||
}
|
||||
// 统计纵向
|
||||
vector<int> vertical(m , 0);
|
||||
for (int j = 0; j < m; j++) {
|
||||
for (int i = 0 ; i < n; i++) {
|
||||
vertical[j] += vec[i][j];
|
||||
}
|
||||
}
|
||||
int result = INT_MAX;
|
||||
int horizontalCut = 0;
|
||||
for (int i = 0 ; i < n; i++) {
|
||||
horizontalCut += horizontal[i];
|
||||
result = min(result, abs(sum - horizontalCut - horizontalCut));
|
||||
}
|
||||
int verticalCut = 0;
|
||||
for (int j = 0; j < m; j++) {
|
||||
verticalCut += vertical[j];
|
||||
result = min(result, abs(sum - verticalCut - verticalCut));
|
||||
}
|
||||
cout << result << endl;
|
||||
}
|
||||
```
|
||||
79
problems/kamacoder/0130.小美的字符串变换.md
Normal file
79
problems/kamacoder/0130.小美的字符串变换.md
Normal file
@ -0,0 +1,79 @@
|
||||
|
||||
# 130.小美的字符串变换
|
||||
|
||||
本题是[岛屿数量](./0099.岛屿的数量广搜.md)的进阶版，主要思路和代码都是一样的，统计一个图里岛屿的数量，也是染色问题。
|
||||
|
||||
1、 先枚举各个可能出现的矩阵
|
||||
2、 针对矩阵经行广搜染色（深搜，并查集一样可以）
|
||||
3、 统计岛屿数量最小的数量。
|
||||
|
||||
```CPP
|
||||
#include <iostream>
|
||||
#include <climits>
|
||||
#include <vector>
|
||||
#include <queue>
|
||||
using namespace std;
|
||||
|
||||
// 广搜代码同 卡码网：99. 岛屿数量
|
||||
int dir[4][2] = {0, 1, 1, 0, -1, 0, 0, -1}; // 四个方向
|
||||
void bfs(const vector<vector<char>>& grid, vector<vector<bool>>& visited, int x, int y, char a) {
|
||||
queue<pair<int, int>> que;
|
||||
que.push({x, y});
|
||||
visited[x][y] = true; // 只要加入队列，立刻标记
|
||||
while(!que.empty()) {
|
||||
pair<int ,int> cur = que.front(); que.pop();
|
||||
int curx = cur.first;
|
||||
int cury = cur.second;
|
||||
for (int i = 0; i < 4; i++) {
|
||||
int nextx = curx + dir[i][0];
|
||||
int nexty = cury + dir[i][1];
|
||||
if (nextx < 0 || nextx >= grid.size() || nexty < 0 || nexty >= grid[0].size()) continue; // 越界了，直接跳过
|
||||
if (!visited[nextx][nexty] && grid[nextx][nexty] == a) {
|
||||
que.push({nextx, nexty});
|
||||
visited[nextx][nexty] = true; // 只要加入队列立刻标记
|
||||
}
|
||||
}
|
||||
}
|
||||
}
|
||||
|
||||
|
||||
int main() {
|
||||
int n;
|
||||
string s;
|
||||
cin >> n;
|
||||
int result = INT_MAX;
|
||||
cin >> s;
|
||||
for (int k = 1; k < n; k++) {
|
||||
if (n % k != 0) continue;
|
||||
// 计算出 矩阵的 行 和 列
|
||||
int x = n / k;
|
||||
int y = k;
|
||||
//cout << x << " " << y << endl;
|
||||
vector<vector<char>> vec(x, vector<char>(y, 0));
|
||||
// 填装矩阵
|
||||
int sCount = 0;
|
||||
for (int i = 0; i < x; i++) {
|
||||
for (int j = 0; j < y; j++) {
|
||||
vec[i][j] = s[sCount++];
|
||||
}
|
||||
}
|
||||
|
||||
// 开始广搜染色
|
||||
vector<vector<bool>> visited(x, vector<bool>(y, false));
|
||||
int count = 0;
|
||||
for (int i = 0; i < x; i++) {
|
||||
for (int j = 0; j < y; j++) {
|
||||
|
||||
if (!visited[i][j]) {
|
||||
count++; // 遇到没访问过的陆地，+1
|
||||
bfs(vec, visited, i, j, vec[i][j]); // 将与其链接的陆地都标记上 true
|
||||
}
|
||||
}
|
||||
}
|
||||
// 取岛屿数量最少的
|
||||
result = min (result, count);
|
||||
|
||||
}
|
||||
cout << result << endl;
|
||||
}
|
||||
```
|
||||
5
problems/kamacoder/0131.小美的树上染色.md
Normal file
5
problems/kamacoder/0131.小美的树上染色.md
Normal file
@ -0,0 +1,5 @@
|
||||
# 131. 小美的树上染色
|
||||
|
||||
贪心的思路 ： https://blog.csdn.net/weixin_43739821/article/details/136299012
|
||||
|
||||
dp思路：https://www.cnblogs.com/ganyq/p/18111114
|
||||
29
problems/kamacoder/小美的排列询问.md
Normal file
29
problems/kamacoder/小美的排列询问.md
Normal file
@ -0,0 +1,29 @@
|
||||
|
||||
小美的排列询问
|
||||
|
||||
注意 x 和y 不分先后
|
||||
|
||||
```CPP
|
||||
|
||||
#include <iostream>
|
||||
#include <vector>
|
||||
using namespace std;
|
||||
int main() {
|
||||
int n, x, y;
|
||||
cin >> n;
|
||||
vector<int> vec(n, 0);
|
||||
for (int i =0; i < n; i++) {
|
||||
cin >> vec[i];
|
||||
}
|
||||
cin >> x >> y;
|
||||
for (int i = 0; i < n - 1; i++) {
|
||||
if (x == vec[i] && y == vec[i + 1]) || (y == vec[i] && x == vec[i + 1]) ) {
|
||||
cout << "Yes" << endl;
|
||||
return 0;
|
||||
}
|
||||
}
|
||||
cout << "No" << endl;
|
||||
|
||||
}
|
||||
|
||||
```
|
||||
30
problems/kamacoder/小美走公路.md
Normal file
30
problems/kamacoder/小美走公路.md
Normal file
@ -0,0 +1,30 @@
|
||||
|
||||
|
||||
两个注意点
|
||||
|
||||
1. x 可以比 y 大，题目没规定 x 和y 的大小顺序
|
||||
2. 累计相加的数可能超过int
|
||||
|
||||
|
||||
```CPP
|
||||
#include <iostream>
|
||||
#include <vector>
|
||||
using namespace std;
|
||||
int main () {
|
||||
int n;
|
||||
cin >> n;
|
||||
vector<int> vec(2* n + 1, 0);
|
||||
for (int i = 1; i <= n; i++) {
|
||||
cin >> vec[i];
|
||||
vec[n + i] = vec[i];
|
||||
}
|
||||
int x, y;
|
||||
cin >> x >> y;
|
||||
int xx = min(x ,y); // 注意点1：x 可以比 y 大
|
||||
int yy = max(x, y);
|
||||
long long a = 0, b = 0; // 注意点2：相加的数可能超过int
|
||||
for (int i = xx; i < yy; i++) a += vec[i];
|
||||
for (int i = yy; i < xx + n; i++ ) b += vec[i];
|
||||
cout << min(a, b) << endl;
|
||||
}
|
||||
```
|
||||
Reference in New Issue
Block a user