标签:动态规划
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
Return 4.
/* * 动态规划的算法 *if(m[i][j] = 1) d[i][j] = min(d[i-1][j-1], d[i][j-1], d[i-1][j]) + 1 *m[i][j] = 0; d[i][j] = 0; *初始化 d[0][j] = m[0][j]; d[i][0] = m[i][0]; *优化思路用一行d[j] 进行 *int preNode = d[0]; *d[0] = m[i][0]; * for(int j = 1; j < n; j++) * { * if(m[i][j] = 1) int temp = min(preNode, d[j-1], d[j]) + 1; * preNode = d[j]; * d[j] = temp; * } *错误1:没有考虑<1,0,1,1>向量 ;把矩阵想成等宽高的 */int min(int a, int b) { return a < b ? a : b; } int min(int a, int b, int c) { return min(min(a,b),min(b,c)); } int maximalSquare(vector<vector<char>>& m) { if(m.size() == 0) return 0; int m_size = m.size(); // if(m_size == 1) return (m[0][0] == '0') ? 0 : 1; int* d = new int[m[0].size()]; int max = 0; //init for(int i = 0; i < m[0].size(); i++) { d[i] = (m[0][i] == '0') ? 0 : 1; if(d[i] > max) max = d[i]; } //循环 for(int l = 1; l < m_size; l++) //从第1行开始 { int preNode = d[0]; d[0] = (m[l][0] == '0') ? 0 : 1; for(int j = 1; j < m[0].size(); j++) { if(m[l][j] == '0') { preNode = d[j]; d[j] = 0; } else //m[l][j] = 1时 { int temp = min(preNode, d[j-1], d[j]);//d[l-1][j-1], d[l][j-1], d[l-1][j] preNode = d[j]; d[j] = temp + 1; if(d[j] > max) { max = d[j]; } } } } delete d; return max*max; }
标签:动态规划
原文地址:http://blog.csdn.net/fy_sun123/article/details/46347547