Largest Rectangle in Histogram leetcode java

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：

        这道题自己是完全没想到用栈了。。

        有个很完整很详细很好的讲解在这里： http://www.cnblogs.com/lichen782/p/leetcode_Largest_Rectangle_in_Histogram.html

        我就不写了。贴一下上面提到的代码吧。

        O(n^2)的：

public int largestRectangleArea(int[] height) {        // Start typing your Java solution below        // DO NOT write main() function        int[] min = new int[height.length];        int maxArea = 0;        for(int i = 0; i < height.length; i++){            if(height[i] != 0 && maxArea/height[i] >= (height.length - i)) {                continue;            }            for(int j = i; j < height.length; j++){                if(i == j) min[j] = height[j];                else {                    if(height[j] < min[j - 1]) {                        min[j] = height[j];                    }else min[j] = min[j-1];                }                int tentativeArea = min[j] * (j - i + 1);                if(tentativeArea > maxArea) {                    maxArea = tentativeArea;                }            }        }        return maxArea;    }

O(n)的： 

public int largestRectangleArea2(int[] height) {        Stack<Integer> stack = new Stack<Integer>();        int i = 0;        int maxArea = 0;        int[] h = new int[height.length + 1];        h = Arrays.copyOf(height, height.length + 1);        while(i < h.length){            if(stack.isEmpty() || h[stack.peek()] <= h[i]){                stack.push(i++);            }else {                int t = stack.pop();                maxArea = Math.max(maxArea, h[t] * (stack.isEmpty() ? i : i - stack.peek() - 1));            }        }        return maxArea;    }