LeetCode: Largest Rectangle in Histogram

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.

地址：https://oj.leetcode.com/problems/largest-rectangle-in-histogram/

算法： 计算以每一个bar的高为宽的最大长方形面积，则面积最大的那个即为所求的最大面积。对于以第i个bar为高的长方形可以这样求，向左找到第一个比这个bar低的bar的下标，记为left_index，向右找到第一个比这个bar低的bar的下标，记为right_index，这个bar对应的长方形面积=h[i]*(right_index-left_index)。利用栈来计算该过程可使时间复杂度达到O(n)。首先，入栈的条件是：空栈或者当前遍历的bar的高度比栈顶的高度高，若达不到上诉条件则出栈。这样就可以保证，当前遍历的节点一定是出栈元素的left_index，而出栈元素的下一个节点为right_index。代码：

class Solution {
public:
    int largestRectangleArea(vector<int> &height) {
        int len = height.size();
        if(len < 1) return 0;
        stack<int> stk;
        int i = 0;
        int max_area = 0;
        while(i < len){
            if(stk.empty() || height[stk.top()] <= height[i]){
                stk.push(i++);
            }else{
                int t = stk.top();
                stk.pop();
                int area = height[t] * (stk.empty() ? i : i - stk.top() - 1);
                if(area > max_area){
                    max_area = area;
                }
            }
        }
        while(!stk.empty()){
            int t = stk.top();
            stk.pop();
            int area = height[t] * (stk.empty() ? len : len - stk.top() - 1);
            if(area > max_area){
                max_area = area;
            }
        }
        return max_area;
    }
};