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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.
代码：
class Solution { public: int largestRectangleArea(vector<int>& height) { int ret = 0; const size_t len = height.size(); if (!len) return len; int *dp_left = new int[len](), *dp_right = new int[len](); dp_left[0] = 0; dp_right[len-1] = len-1; // dp_left : record the left most position of height arr that the curr element can reach for ( size_t i = 1; i < len; ++i ){ int left = i; while ( left>0 && height[i]<=height[left-1] ) left = dp_left[left-1]; dp_left[i] = left; } // dp_right : vice versa for ( int i = len-2; i >0; --i) { int right = i; while ( right<len-1 && height[i]<=height[right+1] ) right = dp_right[right+1]; dp_right[i] = right; } // get the largest rectangle for ( size_t i = 0; i < len; ++i ) ret = std::max( ret, (dp_right[i]-dp_left[i]+1)*height[i] ); delete[] dp_left; delete[] dp_right; return ret; } };
tips：
采用dp的思路，主要参考 http://www.acmerblog.com/largest-rectangle-in-histogram-6117.html
遍历三次：
1. left to right : dp_left[i]数组存放height[i]向左最多能推到哪个位置
2. right to left : dp_right[i]数组存放height[i]向右最多能推到哪个位置
注意，不论是dp_left还是dp_right存放都是height数组的绝对位置（一开始dp_right一直存放相对位置，导致不能AC）
这里虽然for循环中又有while循环，但是复杂度并不是O(n²)，原因并不是一个挨着一个跳的，而是用之前比较的结果(dp_left，dp_right)跳的。
3. 最后走一遍dp，记录最大值
【Largest Rectangle in Histogram】cpp
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原文地址：http://www.cnblogs.com/xbf9xbf/p/4499450.html
