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问题描述
Given n non-negative integers representing the histogram‘s bar height where the width of each bar is 1, find the area of largest rectangle in the histogram.
Above is a histogram where width of each bar is 1, given height = [2,1,5,6,2,3].
The largest rectangle is shown in the shaded area, which has area = 10 unit.
For example,
Given height = [2,1,5,6,2,3],
return 10.
解决思路
巧妙的利用辅助栈。
（1）栈内存储的是数组遍历过程中的非递减元素的下标；
（2）如果遍历的当前元素比栈顶元素更小，则弹出栈顶值（下标）的索引元素，作为高度；宽度要看栈是否为空，如果为空则直接为下标值，否则为当前遍历的下标减去栈顶值再减1；
（3）遍历结束，如果栈内还有元素，则继续计算。
时间/空间复杂度均为O(n).
程序
public class Solution {
public int largestRectangleArea(int[] height) {
if (height == null || height.length == 0) {
return 0;
}
Stack<Integer> s = new Stack<>();
int max = 0;
for (int i = 0; i <= height.length; i++) {
int cur = i == height.length ? -1 : height[i]; // trick
while (!s.isEmpty() && cur < height[s.peek()]) {
int h = height[s.pop()];
int w = s.isEmpty() ? i : i - s.peek() - 1;
max = Math.max(max, h * w);
}
s.push(i);
}
return max;
}
}
Follow up
Maximal Rectangle
问题描述
Given a 2D binary matrix filled with 0‘s and 1‘s, find the largest rectangle containing all ones and return its area.
解决思路
假设矩阵的行数为n, 列数为m.
首先将矩阵转化为n个rectangle，然后利用上面的方法计算max，最后输出这n个中的max即可。
程序
public class Solution {
public int maximalRectangle(char[][] matrix) {
if (matrix == null || matrix.length == 0 || matrix[0].length == 0) {
return 0;
}
int[][] nums = transform(matrix);
int max = 0;
for (int i = 0; i < matrix.length; i++) {
max = Math.max(max, getMaxRectangle(nums[i]));
}
return max;
}
private int getMaxRectangle(int[] nums) {
if (nums == null || nums.length == 0) {
return 0;
}
Stack<Integer> s = new Stack<>();
int max = 0;
for (int i = 0; i <= nums.length; i++) {
int cur = i == nums.length ? -1 : nums[i];
while (!s.isEmpty() && cur < nums[s.peek()]) {
int h = nums[s.pop()];
int w = s.isEmpty() ? i : i - s.peek() - 1;
max = Math.max(max, h * w);
}
s.push(i);
}
return max;
}
// transform to single rectangle
private int[][] transform(char[][] matrix) {
int n = matrix.length, m = matrix[0].length;
int[][] nums = new int[n][m];
// first row
for (int j = 0; j < m; j++) {
nums[0][j] = matrix[0][j] - ‘0‘;
}
//other
for (int i = 1; i < n; i++) {
for (int j = 0; j < m; j++) {
nums[i][j] = matrix[i][j] == ‘0‘ ? 0 : nums[i - 1][j] + 1;
}
}
return nums;
}
}
Largest Rectangle in Histogram
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原文地址：http://www.cnblogs.com/harrygogo/p/4695190.html
