LeetCode200 Number of Islands

LeetCode200 Number of Islands

問題來源 LeetCode200

問題描述

Given a 2d grid map of ‘1’s (land) and ‘0’s (water), count the number of islands. An island is surrounded by water and is formed by connecting adjacent lands horizontally or vertically. You may assume all four edges of the grid are all surrounded by water.

Example 1:

11110
11010
11000
00000

Answer: 1

Example 2:

11000
11000
00100
00011 

Answer: 3

問題分析

這道題我的解法是遞歸的判斷每個爲“1”的方格內的元素的四周是否也爲“1”，很明顯會產生重複判斷造成死循環，所以使用動態規劃，維護一個和char[][]數組相同的數組來記錄哪些爲“1”的點是已經判斷過的。
算法過程如下圖：

這裏需要注意的一點是，因爲題中需要的是獨立的島的個數，所以需要維護一個全局變量，來保存。我們可以認爲一次完成一次遞歸就代表遍歷了一個島嶼，所以每次找到新的小島，就進行累加。完成統計

具體代碼如下

代碼

int res =0;
public int numIslands(char[][] grid) {
    if(grid==null||grid.length<1||grid[0].length<1) return 0;
    int m =grid.length,n = grid[0].length;
    boolean[][] dp = new boolean[m][n];
    for (int i = 0; i <m ; i++) {
        for (int j = 0; j <n ; j++) {
            if(!dp[i][j]&&grid[i][j]=='1'){
                help(grid,dp,i,j);
                res++;
            }
        }
    }
    return res;
}
void help(char[][] grid,boolean[][] dp,int i,int j){
    if(dp[i][j]) return;
    if(grid[i][j]=='1'){
        dp[i][j]=true;
        if(i>0) help(grid,dp,i-1,j);
        if(j>0) help(grid,dp,i,j-1);
        if(i<grid.length-1) help(grid,dp,i+1,j);
        if(j<grid[0].length-1) help(grid,dp,i,j+1);
    }
}

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