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;
     }