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  • LeetCode OJ:Maximal Square(最大矩形)

    Given a 2D binary matrix filled with 0's and 1's, find the largest square containing all 1's and return its area.

    For example, given the following matrix:

    1 0 1 0 0
    1 0 1 1 1
    1 1 1 1 1
    1 0 0 1 0

    在给定的二维字符数组,找出最大全为1的矩形包含的1的个数:

    注意这里相当于求面积而已,只要求出边长就很好办了。边长用dp很好求,这题其实跟前面有道dp问题很像,以后有时间找出来,先贴下代码:

     1 class Solution {
     2 public:
     3     int maximalSquare(vector<vector<char>>& matrix) {
     4         if(!matrix.size()||!matrix[0].size()) return 0;
     5         vector<vector<int>>dp(matrix.size(), vector<int>(matrix[0].size(), 0));
     6         int maxVal = 0;
     7         for(int i = 0; i < matrix.size(); ++i){
     8             if(matrix[i][0] == '1'){
     9                 dp[i][0] = 1;
    10                 maxVal = 1;
    11             }else dp[i][0] = 0;
    12         }
    13         for(int j = 0; j < matrix[0].size(); ++j){
    14             if(matrix[0][j] == '1'){
    15                 dp[0][j] = 1; 
    16                 maxVal = 1;
    17             }else dp[0][j] = 0;
    18         }
    19         for(int i = 1; i < matrix.size(); ++i){
    20             for(int j = 1; j < matrix[0].size(); ++j){
    21                 if(matrix[i][j] == '1'){
    22                     dp[i][j] = min(dp[i-1][j], min(dp[i][j-1], dp[i-1][j-1])) + 1;
    23                     maxVal = max(maxVal, dp[i][j]);
    24                 }else
    25                     dp[i][j] = 0;
    26             }
    27         }
    28         return maxVal*maxVal;
    29     }
    30 };
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  • 原文地址:https://www.cnblogs.com/-wang-cheng/p/4975068.html
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