LeetCode 算法入门 Day7 广度优先搜索 / 深度优先搜索 Java

733. 图像渲染

题目描述:

代码:

 

class Solution {
    int []dx={1,0,0,-1};
    int []dy={0,1,-1,0};
    public int[][] floodFill(int[][] image, int sr, int sc, int newColor) {
        int currColor = image[sr][sc];
        if(currColor==newColor){
            return image;
        }
        int n=image.length,m=image[0].length;
        Queue<int[]> queue = new LinkedList<int[]>();
        queue.offer(new int[] {sr,sc});
        image[sr][sc]=newColor;
        while(!queue.isEmpty()){
            int []cell=queue.poll();
            int x=cell[0],y=cell[1];
            for(int i=0;i<4;i++){
                int mx=x+dx[i],my=y+dy[i];
                if(mx>=0&&mx<n&&my>=0&&my<m&&image[mx][my]==currColor){
                    queue.offer(new int[]{mx,my});
                    image[mx][my]=newColor;
                }
            }
        }
        return image;
    }
}

思路:

 695. 岛屿的最大面积

问题描述:

 代码:

class Solution {
    public int maxAreaOfIsland(int[][] grid) {
      int ans = 0;
      for(int i=0;i!=grid.length;++i){
          for(int j=0;j!=grid[0].length;++j){
              int cur=0;
              Queue<Integer> queuei = new LinkedList<Integer>();
              Queue<Integer> queuej = new LinkedList<Integer>();
              queuei.offer(i);
              queuej.offer(j);
              while(!queuei.isEmpty()){
                  int cur_i = queuei.poll(),cur_j=queuej.poll();
                  if(cur_i<0||cur_j<0||cur_i==grid.length||cur_j==grid[0].length||grid[cur_i][cur_j]!=1){
                      continue;
                  }
                  ++cur;
                  grid[cur_i][cur_j]=0;
                  int[] di = {1,0,0,-1};
                  int[] dj = {0,1,-1,0};

                  for(int index=0;index!=4;index++){
                      int index_x=cur_i+di[index],index_y=cur_j+dj[index];
                      queuei.offer(index_x);
                      queuej.offer(index_y);
                  }
              }
              ans=Math.max(ans,cur);
          }
      }
      return ans;
    }
}

思路: