Leetcode题解84- 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.

 

Example:

Input: [2,1,5,6,2,3]
Output: 10

思路：记录左方向和右方向第一个小于当前值的下标。O(n)复杂度。

for (int i = 1; i < height.length; i++) {              
    int p = i - 1;
    while (p >= 0 && height[p] >= height[i]) {
        p--;
    }
    lessFromLeft[i] = p;              
}

The only line change shifts this algorithm from O(n^2) to O(n) complexity: we don't need to rescan each item to the left - we can reuse results of previous calculations and "jump" through indices in quick manner:

while (p >= 0 && height[p] >= height[i]) {
      p = lessFromLeft[p];
}

代码：

class Solution {
public:
    int largestRectangleArea(vector<int>& heights) {
        if(heights.size() == 0) return 0;
        int Len = heights.size();
        vector<int> lessFromRight(Len),lessFromLeft(Len);
        lessFromRight[Len-1] = Len;
        lessFromLeft[0]=-1;
        for(int i = 1; i < Len; i++){
            int p = i - 1;
            while(p >= 0 && heights[p]>=heights[i]) p = lessFromLeft[p];  //左侧方向第一个比当前值小的下标
            lessFromLeft[i] = p;
        }
        for(int i = Len - 2; i >= 0; i--){
            int p = i + 1;
            while(p < Len && heights[p] >= heights[i]) p = lessFromRight[p]; //右侧方向第一个比当前值小的下标
            lessFromRight[i] = p;
        }
        int maxArea = 0;
        for(int i = 0; i < Len; i++){
            maxArea = max(maxArea, heights[i]*(lessFromRight[i] - lessFromLeft[i] - 1));
        }
        return maxArea;
    }
};