LEETCODE: Largest Rectangle in Histogram

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.

参考： http://www.cnblogs.com/remlostime/archive/2012/11/25/2787359.html

我就做了最土的那种，先从左向右扫描，再从右向做扫描，得到当前值可以跨越几行。然后求最大值。

class Solution {
public:
    int largestRectangleArea(vector<int> &height) {
        vector<int> expand(height.size());
        stack<int> exstack;
        
        // Get the expand from left to right.
        for(int ii = 0; ii < height.size(); ii ++) {
            if(exstack.empty()) {
                expand[ii] = 0;
            }
            else {
                while(!exstack.empty()) {
                    if(height[ii] <= height[exstack.top()]) {
                        exstack.pop();
                    }
                    else {
                        break;
                    }
                }
                if(exstack.empty()) {
                    expand[ii] = ii;
                }
                else {
                    expand[ii] = ii - exstack.top() - 1;
                }
            }
            exstack.push(ii);
        }
        
        while(!exstack.empty())
            exstack.pop();
        
        // Get the expand from right to left.
        for(int ii = height.size() - 1; ii >= 0; ii --) {
            if(exstack.empty()) {
                expand[ii] += 0;
            }
            else {
                while(!exstack.empty()) {
                    if(height[exstack.top()] >= height[ii]) {
                        exstack.pop();
                    }
                    else {
                        break;
                    }
                }
                if(exstack.empty()) {
                    expand[ii] += height.size() - 1 - ii;
                }
                else {
                    expand[ii] += exstack.top() - 1 - ii;
                }
            }
            exstack.push(ii);
        }
        
        // Get the max.
        int maxval = 0;
        for(int ii = 0; ii < height.size(); ii ++) {
            maxval = max(maxval, height[ii] * (expand[ii] + 1));
        }
        
        return maxval;
    }
};