38 - 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!

solution：原本的算法是找出当前点的左右最大值，取最小，减去当前的高度，即得到当前点的容水量，不过需要两个数组来存左右最大值，在论坛上看到一个非常巧妙的算法没有额外空间：其核心还是一样，但是关键之处是按照数组的最大值将输入数组划分成两部分，左边部分小于最大值，右边也是，左边部分循环时就可以不用比较左右的最小值了，只需要不停保存左最大值即可，因为右边存在最大值了；同理引申到右半部分。

需要注意的一点是边界的时候最大值为0.

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(A == NULL || n <= 1)
            return 0;
        
        int res = 0;
        
        int maxIndex = 0;
        for(int i = 0; i < n; i++)
        {
            if(A[i] > A[maxIndex])
                maxIndex = i;
        }
        
        int leftMax = 0;
        for(int i = 0; i < maxIndex; i++)
        {
            if(leftMax > A[i])
                res += leftMax-A[i];
            else
                leftMax = A[i];
        }
        
        int rightMax = 0;
        for(int i = n-1; i > maxIndex; i--)
        {
            if(rightMax > A[i])
                res += rightMax-A[i];
            else
                rightMax = A[i];
        }
        
        return res;
    }
};