Leetcode：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!

刚开始做错了，先找下降的，再找上升的，代码如下

class Solution {
public:
    int trap(int A[], int n) {
        // Note: The Solution object is instantiated only once and is reused by each test case.
         if(NULL==A || n<=1)
            return 0;
        int beg=0,end=1,change=-1,max=0;
        int result = 0;
        while(begA[end]) //is down
            {
                ++end;  
            }
            if(end<=n && end-beg>1)//is down
            {
                max = A[beg];
                change = end-1; 
                //up
                int tmp = 0;
                while(endA[end-1])
                {
                    max = A[end-1]; 
                }
                //down
                for(int j=beg+1; j

但

Input:	[4,2,0,3,2,5]
Output:	5
Expected:	9

错误，后来恍然大悟，如果两头构成最大的显然是错的，真是傻逼了。。。

分析：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。时间复杂度是O(n)。

int trap(int A[], int n) {
        // Note: The Solution object is instantiated only once.
        if(A==NULL || n<1)return 0;
    	
		int maxheight = 0;
		vector leftMostHeight(n);
		for(int i =0; i A[i] ? maxheight : A[i];
		}

		maxheight = 0;
		vector rightMostHeight(n);
		for(int i =n-1;i>=0;i--)
		{
			rightMostHeight[i] = maxheight;
			maxheight = maxheight > A[i] ? maxheight : A[i];
		}

		int water = 0;
		for(int i =0; i < n; i++)
		{
			int high = min(leftMostHeight[i],rightMostHeight[i])-A[i];
			if(high>0)
				water += high;
		}
		return water;
    }