标签:rectangle empty als str 直接 http htm math char
其实这题是85. 最大矩形的特殊情况,我们将85题代码稍微改一下,然后直接套用即可。
此题要求是正方形,那么我们在计算长与宽时,取短的那条然后平方即可。
class Solution {
public int maximalSquare(char[][] matrix) {
return maximalRectangle(matrix);
}
public int maximalRectangle(char[][] matrix) {
int m = matrix.length;
if (m == 0) return 0;
int n = matrix[0].length;
if (n == 0) return 0;
int[] height = new int[n];
int ans = 0;
for (int i = 0; i < m; i++) {
for (int j = 0; j < n; j++) {
if (matrix[i][j] == ‘0‘) height[j] = 0;
else if (matrix[i][j] == ‘1‘) height[j] += 1;
}
ans = Math.max(largestRectangleArea(height), ans);
}
return ans;
}
public int largestRectangleArea(int[] heights) {
if (heights.length == 0) return 0;
Stack<Integer> stack = new Stack<>();
int max = 0;
for (int i = 0; i < heights.length; i++) {
while (!stack.isEmpty() && heights[stack.peek()] >= heights[i]) {
int tmp = stack.pop();
// 把当前的tmp木板作为最短木板,看能组成的最大面积是多少
int bian = Math.min(heights[tmp], (stack.empty() ? i : i - stack.peek() - 1));
// max = Math.max(max, heights[tmp] * (stack.empty() ? i : i - stack.peek() - 1));
max = Math.max(max, bian * bian);
}
stack.push(i);
}
int tmp = 0;
int len = heights.length;
while (!stack.isEmpty()) {
tmp = stack.pop();
int bian = Math.min(heights[tmp], (stack.empty() ? len : len - stack.peek() - 1));
// max = Math.max(max, heights[tmp] * (stack.empty() ? len : len - stack.peek() - 1));
max = Math.max(max, bian * bian);
}
return max;
}
}
标签:rectangle empty als str 直接 http htm math char
原文地址:https://www.cnblogs.com/acbingo/p/9404167.html