LeetCode 题解（99）: Largest Rectangle in Histogram

题目：

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.

题解：

网上学来神算法。用堆栈保存height的index。

C++版：

class Solution:
    # @param {integer[]} height
    # @return {integer}
    def largestRectangleArea(self, height):
        height.append(0)
        maxArea = 0
        i = 0
        index = []
        while i < len(height):
            if len(index) == 0 or height[index[-1]] <= height[i]:
                index.append(i)
                i += 1
            else:
                t = index.pop()
                if len(index) == 0:
                    maxArea = max(maxArea, height[t] * i)
                else:
                    maxArea = max(maxArea, height[t] * (i - 1 - index[-1]))
        return maxArea

Java版：

public class Solution {
    public int largestRectangleArea(int[] height) {
        int[] h = new int[height.length + 1];
        h = Arrays.copyOf(height, height.length + 1);
        Stack<Integer> index = new Stack<>();
        int i = 0;
        int maxArea = 0;
        while(i < h.length) {
            if(index.isEmpty() || h[index.peek()] <= h[i]) {
                index.push(i++);
            } else {
                int t = index.pop();
                maxArea = Math.max(maxArea, h[t] * (index.isEmpty() ? i : i - 1 - index.peek()));
            }
        }
        return maxArea;
    }
}

Python版：

class Solution:
    # @param {integer[]} height
    # @return {integer}
    def largestRectangleArea(self, height):
        height.append(0)
        maxArea = 0
        i = 0
        index = []
        while i < len(height):
            if len(index) == 0 or height[index[-1]] <= height[i]:
                index.append(i)
                i += 1
            else:
                t = index.pop()
                if len(index) == 0:
                    maxArea = max(maxArea, height[t] * i)
                else:
                    maxArea = max(maxArea, height[t] * (i - 1 - index[-1]))
        return maxArea