LEECODE：DFS算法（104、494、547、37）——java实现

LeetCode104.Maximum Depth of Binary Tree 计算最大深度。

比较经典的递归做法。

//找到二叉树的最大深度
//离根节点最远的叶子节点
/**
 * Definition for a binary tree node.
 * public class TreeNode {
 *     int val;
 *     TreeNode left;
 *     TreeNode right;
 *     TreeNode(int x) { val = x; }
 * }
 */
class Solution {
    public int maxDepth(TreeNode root) {
        if(root==null) return 0;
        else{//m为左侧的深度，n为右侧的深度
             //递归，深搜
            int m=maxDepth(root.left);
            int n=maxDepth(root.right);
            if(m>n) return m+1;
            else return n+1;
        }
    }
}

LeetCode494 TargetSum 使用dfs算法来做，比较简单直接。

//深度搜索从数组开始进行求和。

You are given a list of non-negative integers, a1, a2, ..., an, and a target, S. Now you have 2 symbols + and -. For each integer, you should choose one from + and - as its new symbol.

Find out how many ways to assign symbols to make sum of integers equal to target S.

Example 1:

Input: nums is [1, 1, 1, 1, 1], S is 3. 
Output: 5
Explanation: 

-1+1+1+1+1 = 3
+1-1+1+1+1 = 3
+1+1-1+1+1 = 3
+1+1+1-1+1 = 3
+1+1+1+1-1 = 3

There are 5 ways to assign symbols to make the sum of nums be target 3.

class Solution {
    public int findTargetSumWays(int[] nums, int S) {
        return  dfs(nums,0,S);
    }
    public int dfs(int[] nums,int i,int sum){
        if(i==nums.length){
            if(sum==0) return 1;//成功
            else return 0;//失败
        }
        int ways=0;
        ways+=dfs(nums,i+1,sum-nums[i]); //-号遍历
        ways+=dfs(nums,i+1,sum+nums[i]); //+号遍历
        return ways;//返回结果
    }
}

LeetCode547 FriendCircle（朋友圈）

//计算朋友圈的数量
//题目中的意思是第一个朋友圈中有((0，0),(0，1),(1，0),(1，1))，第二个朋友圈是(2，2)
//深度优先遍历能遍历所有值为1的结点，通过图的连通性来实现。
//visit记录连通性的好坏。
public class Solution {
    public int findCircleNum(int[][] M) {
        int[] visited = new int[M.length];
        int count = 0;
        for (int i = 0; i < M.length; i++) {
            if (visited[i] == 0) {
                dfs(M, visited, i);
                count++;
            }
        }
        return count;
    }
    public void dfs(int[][] M, int[] visited, int i) {
        for (int j = 0; j < M.length; j++) {
            if (M[i][j] == 1 && visited[j] == 0) {
                visited[j] = 1;
                dfs(M, visited, j);
            }
        }
    }
    
}

LeetCode37 Sudoku Solver

Write a program to solve a Sudoku puzzle by filling the empty cells.

Empty cells are indicated by the character '.'.

You may assume that there will be only one unique solution.

A sudoku puzzle...

...and its solution numbers marked in red.

//数独游戏本身的玩法是每一个行每一列都为1-9不重复
//循环处理子问题，对于每个格子，带入不同的9个数
//然后判断合法性，如果成立就递归继续，结束后把数字设为“.”
class Solution {
    public void solveSudoku(char[][] board) {
        if(board==null||board.length==0) return;
        dfs(board);
    }
    private boolean dfs(char[][] board){
        for(int i=0;i<board.length;i++){
            for(int j=0;j<board[0].length;j++){
                if(board[i][j]=='.'){
                    for(char c='1';c<='9';c++){
                        if(isValid(board,i,j,c)){
                            board[i][j]=c;
                        if(dfs(board)) return true;
                        else board[i][j]='.';
                    }
                }
                /*运行到此说明这个位置不能确定具体的数字*/
                return false;
            }
        }
        }
        return true;
    }
    //判断是否是一个正确的数独（36）
    private boolean isValid(char[][] board,int row,int col,char c){
        for(int i=0;i<board.length;i++){
            //判断col列和row行是否满足数独条件
            if(board[i][col]==c||board[row][i]==c) return false;
            //判断九宫格
            if(board[3*(row/3)+i/3][3*(col/3)+i%3]==c)
                return false;
        }
        return true;
    }
}

leetcode37题参考为：http://blog.youkuaiyun.com/mine_song/article/details/70195463

题解思路有参考。