Trapping Rain Water

Trapping Rain Water
Given n non-negative integers representing an elevation map where the width of each bar is 1, compute how much water it is able to trap after raining.
For example,

Given [0,1,0,2,1,0,1,3,2,1,2,1], return 6.
The above elevation map is represented by array [0,1,0,2,1,0,1,3,2,1,2,1]. In this case, 6 units of rain water (blue section) are being trapped. Thanks Marcos for contributing this image!

算法很简单，核心思想是：对某个值A[i]来说，能trapped的最多的water取决于在i之前最高的值leftMostHeight[i]和在i右边的最高的值rightMostHeight[i]。（均不包含自身）。如果min(left,right) > A[i]，那么在i这个位置上能trapped的water就是min(left,right) – A[i]。

有了这个想法就好办了，第一遍从左到右计算数组leftMostHeight，第二遍从右到左计算rightMostHeight，在第二遍的同时就可以计算出i位置的结果了，而且并不需要开空间来存放rightMostHeight数组。

时间复杂度是O(n)，只扫了两遍。

public class Solution {
    public int trap(int[] A) {
        
        int n = A.length;
        //no way to contain any water
        if(n <= 2) return 0;
        
        //1. first run to calculate the heiest bar on the left of each bar
        int[] lmh =new int[n]; //for the most height on the left
        lmh[0] = 0;
        int maxh =A[0]; // max height
        for(int i =1; i<n; i++){
            lmh[i] = maxh;
            if(A[i] > maxh) maxh = A[i];
        }
        
        int trapped = 0;
        
        
        //2. second run from right to left, 
        // caculate the highest bar on the right of each bar
        // and calculate the trapped water simutaniously
        maxh = A[n-1];
        for(int i = n-2; i>0; i--){
            int left = lmh[i];
            int right = maxh;
            //计算lmh 和 rmh 的最小值
            int container = Math.min(left, right);
            //如果当前值比 container 小，则能容水
            if(container > A[i]){
                trapped += container - A[i];
            }
            
            if(maxh < A[i]) maxh = A[i];
        }
        
        return trapped;
    }
}