leetcode -- Largest Rectangle in Histogram TODO O(N)

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.

本题思路和Trapping Rain Water差不多，计算每个idx的left bound 和right bound（两个bound都需大于height[idx]）

时间复杂度为：O(n^2)

public int largestRectangleArea(int[] height) {
        // Start typing your Java solution below
        // DO NOT write main() function
        int len = height.length;
        int[] area = new int[len];
        
        for(int i = 0; i < len; i++){
            int h = height[i];
            
            int m = i - 1;
            for(; m >= 0; m --){
                if(height[m] < h){
                    break;
                }
            }
            m ++;
            
            int n = i + 1;
            for(; n < len; n ++){
                if(height[n] < h){
                    break;
                }
            }
            n --;
            
            int width = (n - m) + 1;
            area[i] = h * width;
        }
        int max = 0;
        for(int i = 0; i < len; i++){
            if(area[i] > max){
                max = area[i];
            }
        }
        return max;
    }

可以过小数据，大数据挂在输入[1,1,1,1,1,1........] ，说明有很多重复计算，做了一个简单改进，当前高度与上一个相同时，直接将area[i] = area[i-1]（line 12-15）

public int largestRectangleArea(int[] height) {
        // Start typing your Java solution below
        // DO NOT write main() function
        int len = height.length;
        int[] area = new int[len];
        
        for(int i = 0; i < len; i++){
            int h = height[i];
            
12             if(i >= 1 && h == height[i - 1]){
13                 area[i] = area[i - 1];                
14                 continue;
15             }
            
            int m = i - 1;
            for(; m >= 0; m --){
                if(height[m] < h){
                    break;
                }
            }
            m ++;
            
            int n = i + 1;
            for(; n < len; n ++){
                if(height[n] < h){
                    break;
                }
            }
            n --;
            
            int width = (n - m) + 1;
            area[i] = h * width;
        }
        int max = 0;
        for(int i = 0; i < len; i++){
            if(area[i] > max){
                max = area[i];
            }
        }
        return max;
    }