algorithm/leetcode-master
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youngyangyang04
2020-11-17 17:12:33 +08:00
parent db295f4fca
commit c01adfa6bc
5 changed files with 24 additions and 22 deletions
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2
problems/0051.N皇后.md
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@ -89,7 +89,7 @@ void backtracking(int n, int row, vector<string>& chessboard) {
* 递归终止条件
在如下树形结构中：
![51.N皇后](https://img-blog.csdnimg.cn/20201116173551789.png)
![51.N皇后](https://img-blog.csdnimg.cn/20201117165155239.png)
可以看出，当递归到棋盘最底层（也就是叶子节点）的时候，就可以收集结果并返回了。

8
problems/0052.N皇后II.md
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@ -36,7 +36,8 @@ n 皇后问题研究的是如何将 n 个皇后放置在 n×n 的棋盘上，并
class Solution {
private:
int count = 0;
void backtracking(int n, int row, vector<string>& chessboard, vector<vector<string>>& result) {
vector<vector<string>> result;
void backtracking(int n, int row, vector<string>& chessboard) {
if (row == n) {
count++;
return;
@ -44,7 +45,7 @@ void backtracking(int n, int row, vector<string>& chessboard, vector<vector<stri
for (int col = 0; col < n; col++) {
if (isValid(row, col, chessboard, n)) {
chessboard[row][col] = 'Q'; // 放置皇后
backtracking(n, row + 1, chessboard, result);
backtracking(n, row + 1, chessboard);
chessboard[row][col] = '.'; // 回溯
}
}
@ -74,9 +75,10 @@ bool isValid(int row, int col, vector<string>& chessboard, int n) {
public:
int totalNQueens(int n) {
result.clear();
std::vector<std::string> chessboard(n, std::string(n, '.'));
vector<vector<string>> result;
backtracking(n, 0, chessboard, result);
backtracking(n, 0, chessboard);
return count;
}

27
problems/0135.分发糖果.md
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@ -11,7 +11,7 @@
代码如下：
```
```C++
// 从前向后
for (int i = 1; i < ratings.size(); i++) {
if (ratings[i] > ratings[i - 1]) candyVec[i] = candyVec[i - 1] + 1;
@ -37,7 +37,7 @@ for (int i = 1; i < ratings.size(); i++) {
所以代码如下：
```
```C++
// 从后向前
for (int i = ratings.size() - 2; i >= 0; i--) {
if (ratings[i] > ratings[i + 1] ) {
@ -46,21 +46,8 @@ for (int i = ratings.size() - 2; i >= 0; i--) {
}
```
* 将问题分解为若干个子问题
* 找出适合的贪心策略
* 求解每一个子问题的最优解
* 将局部最优解堆叠成全局最优解
* 分解为子问题
这道题目上来也是没什么思路啊
这道题目不好想啊，贪心很巧妙
```
整体代码如下：
```C++
class Solution {
public:
int candy(vector<int>& ratings) {
@ -81,5 +68,9 @@ public:
return result;
}
};
```
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