hot100 | 九、图论

1-leetcode200. 岛屿数量

注意：×

1. 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

2. 注意dfs里面的范围判断，[0, **length-1]**

3. length-1

4. length-1

5. 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：×

1. 没写，看着考的太少，懒得写了

3-leetcode207. 课程表

注意：××

1. 花了很多时间，注意的地方太多了

2. 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<>();

3. 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：×

1. 看了题解再写，感觉还是有点压力

2. 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;
        }
    }