hot100 | 九、图论

1-leetcode200. 岛屿数量

注意：×

蛮巧妙的做法，直接在读取到1的时候给res的值+1,然后深度优先搜索把所有相邻的陆地全部改为海洋
注意dfs里面的范围判断，[0, **length-1]**
length-1
length-1

length-1

 public int numIslands(char[][] grid) {
     int res = 0;
     int n = grid.length;;
     int m = grid[0].length;

     for (int i = 0; i < n; i++) {
         for (int j = 0; j < m; j++) {
             if (grid[i][j] == '1'){
                 res++;
                 dfs(grid, i, j);
             }
         }
     }
     return res;
 }

 private void dfs(char[][] grid, int i, int j) {
     if (i<0 || i>=grid.length || j<0 || j>=grid[0].length){
         return;
     }
     if (grid[i][j] == '0'){
         return;
     }
     grid[i][j] = '0';
     dfs(grid, i+1, j);
     dfs(grid, i, j+1);
     dfs(grid, i-1, j);
     dfs(grid, i, j-1);
 }

leetcode994. 腐烂的橘子

注意：×

没写，看着考的太少，懒得写了

复制代码

3-leetcode207. 课程表

注意：××

花了很多时间，注意的地方太多了
创建List<Integer>[] graph的时候，Labuladong说写代码的时候大部分都是这种邻接表的格式，先创建一个LinkedList类型的数组，然后一定要注意，是graph[i] = new LinkedList<>();

第二个注意的地方是有hasVisited和onPath两种记录，onPath的记录我能理解，hasVisited的记录还需要思考一下，主要的目的就是解决重复搜索

 boolean hasCycle = false;
 boolean[] hasVisited;
 boolean[] onPath;

 public boolean canFinish(int numCourses, int[][] prerequisites) {
     List<Integer>[] graph = buildGraph(numCourses, prerequisites);
     hasVisited = new boolean[numCourses];
     onPath = new boolean[numCourses];

     for (int i = 0; i < numCourses; i++) {
         traverse(graph, i);
     }

     return !hasCycle;
 }

 private void traverse(List<Integer>[] graph, int i) {
     if (onPath[i]){
         hasCycle = true;
     }
     if (hasCycle || hasVisited[i]){
         return;
     }
     
     hasVisited[i] = true;
     
     onPath[i] = true;

     for (Integer integer : graph[i]) {
         traverse(graph, integer);
     }
     
     onPath[i] = false;
 }

 private List<Integer>[] buildGraph(int numCourses, int[][] prerequisites) {
     List<Integer>[] graph = new LinkedList[numCourses];
     for (int i = 0; i < graph.length; i++) {
         graph[i] = new LinkedList<>();
     }
     for (int[] prerequisite : prerequisites) {
         int from = prerequisite[1];
         int to = prerequisite[0];
         graph[from].add(to);
     }
     return graph;
 }

4-leetcode208. 实现 Trie (前缀树)

注意：×

看了题解再写，感觉还是有点压力

注意isEnd的调用

 class Trie {
     class TrieNode{
         boolean isEnd;
         TrieNode[] nodes;
         public TrieNode(){
             isEnd = false;
             nodes = new TrieNode[26];
         }
     }

     private TrieNode root;

     public Trie() {
         root = new TrieNode();
     }

     public void insert(String word) {
         TrieNode node = root;
         for (char c : word.toCharArray()) {
             if (node.nodes[c-'a'] == null){
                 node.nodes[c-'a'] = new TrieNode();
             }
             node = node.nodes[c-'a'];
         }
         node.isEnd = true;
     }

     public boolean search(String word) {
         TrieNode node = root;
         for (char c : word.toCharArray()) {
             if (node.nodes[c-'a'] == null){
                 return false;
             }
             node = node.nodes[c-'a'];
         }
         return node.isEnd;
     }

     public boolean startsWith(String prefix) {
         TrieNode node = root;
         for (char c : prefix.toCharArray()) {
             if (node.nodes[c-'a'] == null){
                 return false;
             }
             node = node.nodes[c-'a'];
         }
         return true;
     }
 }