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https://github.com/youngyangyang04/leetcode-master.git
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Merge branch 'master' of github.com:youngyangyang04/leetcode-master
This commit is contained in:
programmercarl
@ -268,6 +268,34 @@ class Solution {
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}
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```
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```java
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class Solution {
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/**
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* 递归法（思路来自二叉树最大深度的递归法）
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* 该题求最小深度，最小深度为根节点到叶子节点的深度，所以在迭代到每个叶子节点时更新最小值。
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*/
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int depth = 0;
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// 定义最小深度，初始化最大值
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int minDepth = Integer.MAX_VALUE;
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public int minDepth(TreeNode root) {
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dep(root);
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return minDepth == Integer.MAX_VALUE ? 0 : minDepth;
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}
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void dep(TreeNode root){
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if(root == null) return ;
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// 递归开始，深度增加
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depth++;
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dep(root.left);
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dep(root.right);
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// 该位置表示递归到叶子节点了，需要更新最小深度minDepth
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if(root.left == null && root.right == null)
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minDepth = Math.min(minDepth , depth);
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// 递归结束，深度减小
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depth--;
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}
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}
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```
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```Java
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class Solution {
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/**
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@ -652,62 +652,100 @@ function findItinerary(tickets: string[][]): string[] {
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### C
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```C
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char **result;
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bool *used;
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int g_found;
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typedef struct {
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char *name; /* key */
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int cnt; /* 记录到达机场是否飞过了 */
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UT_hash_handle hh; /* makes this structure hashable */
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} to_airport_t;
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int cmp(const void *str1, const void *str2)
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{
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const char **tmp1 = *(char**)str1;
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const char **tmp2 = *(char**)str2;
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int ret = strcmp(tmp1[0], tmp2[0]);
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if (ret == 0) {
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return strcmp(tmp1[1], tmp2[1]);
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typedef struct {
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char *name; /* key */
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to_airport_t *to_airports;
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UT_hash_handle hh; /* makes this structure hashable */
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} from_airport_t;
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void to_airport_destroy(to_airport_t *airports) {
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to_airport_t *airport, *tmp;
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HASH_ITER(hh, airports, airport, tmp) {
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HASH_DEL(airports, airport);
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free(airport);
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}
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return ret;
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}
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void backtracting(char *** tickets, int ticketsSize, int* returnSize, char *start, char **result, bool *used)
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{
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if (*returnSize == ticketsSize + 1) {
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g_found = 1;
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return;
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void from_airport_destroy(from_airport_t *airports) {
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from_airport_t *airport, *tmp;
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HASH_ITER(hh, airports, airport, tmp) {
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to_airport_destroy(airport->to_airports);
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HASH_DEL(airports, airport);
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free(airport);
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}
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}
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int name_sort(to_airport_t *a, to_airport_t *b) {
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return strcmp(a->name, b->name);
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}
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bool backtracking(from_airport_t *airports, int target_path_len, char **path,
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int path_len) {
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if (path_len == target_path_len) return true;
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from_airport_t *from_airport = NULL;
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HASH_FIND_STR(airports, path[path_len - 1], from_airport);
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if (!from_airport) return false;
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for (to_airport_t *to_airport = from_airport->to_airports;
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to_airport != NULL; to_airport = to_airport->hh.next) {
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if (to_airport->cnt == 0) continue;
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to_airport->cnt--;
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path[path_len] = to_airport->name;
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if (backtracking(airports, target_path_len, path, path_len + 1))
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return true;
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to_airport->cnt++;
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}
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return false;
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}
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char **findItinerary(char ***tickets, int ticketsSize, int *ticketsColSize,
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int *returnSize) {
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from_airport_t *airports = NULL;
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// 记录映射关系
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for (int i = 0; i < ticketsSize; i++) {
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if ((used[i] == false) && (strcmp(start, tickets[i][0]) == 0)) {
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result[*returnSize] = (char*)malloc(sizeof(char) * 4);
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memcpy(result[*returnSize], tickets[i][1], sizeof(char) * 4);
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(*returnSize)++;
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used[i] = true;
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/*if ((*returnSize) == ticketsSize + 1) {
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return;
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}*/
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backtracting(tickets, ticketsSize, returnSize, tickets[i][1], result, used);
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if (g_found) {
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return;
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}
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(*returnSize)--;
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used[i] = false;
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from_airport_t *from_airport = NULL;
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to_airport_t *to_airport = NULL;
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HASH_FIND_STR(airports, tickets[i][0], from_airport);
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if (!from_airport) {
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from_airport = malloc(sizeof(from_airport_t));
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from_airport->name = tickets[i][0];
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from_airport->to_airports = NULL;
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HASH_ADD_KEYPTR(hh, airports, from_airport->name,
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strlen(from_airport->name), from_airport);
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}
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HASH_FIND_STR(from_airport->to_airports, tickets[i][1], to_airport);
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if (!to_airport) {
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to_airport = malloc(sizeof(to_airport_t));
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to_airport->name = tickets[i][1];
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to_airport->cnt = 0;
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HASH_ADD_KEYPTR(hh, from_airport->to_airports, to_airport->name,
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strlen(to_airport->name), to_airport);
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}
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to_airport->cnt++;
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}
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return;
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}
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char ** findItinerary(char *** tickets, int ticketsSize, int* ticketsColSize, int* returnSize){
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if (tickets == NULL || ticketsSize <= 0) {
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return NULL;
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// 机场排序
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for (from_airport *from_airport = airports; from_airport != NULL;
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from_airport = from_airport->hh.next) {
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HASH_SRT(hh, from_airport->to_airports, name_sort);
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}
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result = malloc(sizeof(char*) * (ticketsSize + 1));
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used = malloc(sizeof(bool) * ticketsSize);
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memset(used, false, sizeof(bool) * ticketsSize);
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result[0] = malloc(sizeof(char) * 4);
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memcpy(result[0], "JFK", sizeof(char) * 4);
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g_found = 0;
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*returnSize = 1;
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qsort(tickets, ticketsSize, sizeof(tickets[0]), cmp);
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backtracting(tickets, ticketsSize, returnSize, "JFK", result, used);
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char **path = malloc(sizeof(char *) * (ticketsSize + 1));
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path[0] = "JFK"; // 起始机场
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backtracking(airports, ticketsSize + 1, path, 1);
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from_airport_destroy(airports);
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*returnSize = ticketsSize + 1;
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return result;
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return path;
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}
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```
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@ -193,7 +193,7 @@ class Solution {
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class Solution {
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public int[] topKFrequent(int[] nums, int k) {
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// 优先级队列，为了避免复杂 api 操作，pq 存储数组
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// lambda 表达式设置优先级队列从大到小存储 o1 - o2 为从大到小，o2 - o1 反之
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// lambda 表达式设置优先级队列从大到小存储 o1 - o2 为从小到大，o2 - o1 反之
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PriorityQueue<int[]> pq = new PriorityQueue<>((o1, o2) -> o1[1] - o2[1]);
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int[] res = new int[k]; // 答案数组为 k 个元素
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Map<Integer, Integer> map = new HashMap<>(); // 记录元素出现次数
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@ -204,6 +204,7 @@ class Solution {
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tmp[0] = x.getKey();
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tmp[1] = x.getValue();
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pq.offer(tmp);
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// 下面的代码是根据小根堆实现的，我只保留优先队列的最后的k个，只要超出了k我就将最小的弹出，剩余的k个就是答案
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if(pq.size() > k) {
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pq.poll();
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}
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@ -135,7 +135,64 @@ public:
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};
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```
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## 其他语言版本
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### Java：
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```java
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class Solution {
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int[] father;
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public boolean validPath(int n, int[][] edges, int source, int destination) {
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father = new int[n];
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init();
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for (int i = 0; i < edges.length; i++) {
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join(edges[i][0], edges[i][1]);
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}
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return isSame(source, destination);
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}
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// 并查集初始化
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public void init() {
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for (int i = 0; i < father.length; i++) {
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father[i] = i;
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}
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}
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// 并查集里寻根的过程
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public int find(int u) {
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if (u == father[u]) {
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return u;
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} else {
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father[u] = find(father[u]);
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return father[u];
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}
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}
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// 判断 u 和 v是否找到同一个根
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public boolean isSame(int u, int v) {
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u = find(u);
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v = find(v);
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return u == v;
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}
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// 将v->u 这条边加入并查集
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public void join(int u, int v) {
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u = find(u); // 寻找u的根
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v = find(v); // 寻找v的根
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if (u == v) return; // 如果发现根相同，则说明在一个集合，不用两个节点相连直接返回
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father[v] = u;
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}
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}
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```
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### Python：
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PYTHON并查集解法如下：
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```PYTHON
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class Solution:
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def validPath(self, n: int, edges: List[List[int]], source: int, destination: int) -> bool:
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@ -154,4 +211,3 @@ class Solution:
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<a href="https://programmercarl.com/other/kstar.html" target="_blank">
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<img src="../pics/网站星球宣传海报.jpg" width="1000"/>
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</a>
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@ -110,6 +110,9 @@ if (!visited[nextx][nexty] && grid[nextx][nexty] == '1') { // 如果节点没被
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```
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就可以通过 [200.岛屿数量](https://leetcode.cn/problems/number-of-islands/solution/by-carlsun-2-n72a/) 这道题目，大家可以去体验一下。
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## 总结
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当然广搜还有很多细节需要注意的地方，后面我会针对广搜的题目还做针对性的讲解，因为在理论篇讲太多细节，可能会让刚学广搜的录友们越看越懵，所以细节方面针对具体题目在做讲解。
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@ -122,6 +125,37 @@ if (!visited[nextx][nexty] && grid[nextx][nexty] == '1') { // 如果节点没被
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相信看完本篇，大家会对广搜有一个基础性的认识，后面再来做对应的题目就会得心应手一些。
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## 其他语言版本
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### Python
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```python
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from collections import deque
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dir = [(0, 1), (1, 0), (-1, 0), (0, -1)] # 创建方向元素
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def bfs(grid, visited, x, y):
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queue = deque() # 初始化队列
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queue.append((x, y)) # 放入第一个元素/起点
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visited[x][y] = True # 标记为访问过的节点
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while queue: # 遍历队列里的元素
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curx, cury = queue.popleft() # 取出第一个元素
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for dx, dy in dir: # 遍历四个方向
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nextx, nexty = curx + dx, cury + dy
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if nextx < 0 or nextx >= len(grid) or nexty < 0 or nexty >= len(grid[0]): # 越界了，直接跳过
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continue
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if not visited[nextx][nexty]: # 如果节点没被访问过
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queue.append((nextx, nexty)) # 加入队列
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visited[nextx][nexty] = True # 标记为访问过的节点
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```
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<p align="center">
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<a href="https://programmercarl.com/other/kstar.html" target="_blank">
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@ -25,8 +25,8 @@
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这个问题有两个陷阱：
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* 陷阱1：栈是容器适配器，底层容器使用不同的容器，导致栈内数据在内存中是不是连续分布。
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* 陷阱2：缺省情况下，默认底层容器是deque，那么deque的在内存中的数据分布是什么样的呢？ 答案是：不连续的，下文也会提到deque。
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* 陷阱1：栈是容器适配器，底层容器使用不同的容器，导致栈内数据在内存中不一定是连续分布的。
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* 陷阱2：缺省情况下，默认底层容器是deque，那么deque在内存中的数据分布是什么样的呢？ 答案是：不连续的，下文也会提到deque。
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所以这就是考察候选者基础知识扎不扎实的好问题。
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@ -42,7 +42,7 @@
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### 栈在系统中的应用
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如果还记得编译原理的话，编译器在 词法分析的过程中处理括号、花括号等这个符号的逻辑，就是使用了栈这种数据结构。
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如果还记得编译原理的话，编译器在词法分析的过程中处理括号、花括号等这个符号的逻辑，就是使用了栈这种数据结构。
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再举个例子，linux系统中，cd这个进入目录的命令我们应该再熟悉不过了。
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@ -70,8 +70,8 @@ cd a/b/c/../../
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先来分析一下 这里有三种不匹配的情况，
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1. 第一种情况，字符串里左方向的括号多余了 ，所以不匹配。
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2. 第二种情况，括号没有多余，但是 括号的类型没有匹配上。
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1. 第一种情况，字符串里左方向的括号多余了，所以不匹配。
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2. 第二种情况，括号没有多余，但是括号的类型没有匹配上。
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3. 第三种情况，字符串里右方向的括号多余了，所以不匹配。
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这里还有一些技巧，在匹配左括号的时候，右括号先入栈，就只需要比较当前元素和栈顶相不相等就可以了，比左括号先入栈代码实现要简单的多了！
|
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|
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Reference in New Issue
Block a user