leetcode 85 最大矩形(动态规划做法)
作者:互联网
动态规划
很神奇的一个做法。执行速度极快,超越了94%的用户。未完全理解,我先去搞懂84题再回来看 = =。
class Solution {
public:
int maximalRectangle(vector<vector<char>>& matrix) {
if (matrix.size() == 0) return 0;
int leftMax[matrix[0].size()];
fill(leftMax, leftMax + matrix[0].size(), -1);
int rightMin[matrix[0].size()];
fill(rightMin, rightMin + matrix[0].size(), matrix[0].size());
int maxS = 0, boundLeft = -1, boundRight = matrix[0].size();
vector<int> heights(matrix[0].size());
for (int i = 0; i < matrix.size(); i ++) {
for (int j = 0; j < matrix[0].size(); j ++) {
if (matrix[i][j] == '1') {
heights[j] += 1;
} else {
heights[j] = 0;
}
}
boundLeft = -1;
for (int j = 0; j < matrix[0].size(); j ++) {
if (matrix[i][j] == '1') {
leftMax[j] = max(leftMax[j], boundLeft);
} else {
leftMax[j] = -1; //高度为0的矩形左边界为-1
boundLeft = j;
}
}
boundRight = matrix[0].size();
for (int j = matrix[0].size() - 1; j >= 0; j --) {
if (matrix[i][j] == '1') {
rightMin[j] = min(rightMin[j], boundRight);
} else {
rightMin[j] = matrix[0].size();
boundRight = j;
}
}
for (int i = 0; i < matrix[0].size(); i ++) {
maxS = max(maxS, heights[i] * (rightMin[i] - leftMax[i] - 1));
}
}
return maxS;
}
};
标签:leftMax,matrix,int,maxS,rightMin,矩形,85,leetcode,size 来源: https://blog.csdn.net/qq_52993545/article/details/111812770