Trapping Rain Water

Trapping Rain WaterMar 10 '12

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!

对于任意i，决定此列能装多少水的是它左边最大的高度和右边最大的高度。所以，先扫一遍可以得到每个i左边最高的高度值，然后从右向左扫一遍可以得到i右边最高的高度值，这次扫描时可以直接求出i的储水值，所以不再需要存储，直接通过这次扫描可以得到最后的返回值。

代码如下：

class Solution {public:    int trap(int A[], int n) {        // Start typing your C/C++ solution below        // DO NOT write int main() function        if(n <= 2) return 0;                int* left_highest = new int[n];        memset(left_highest,0,sizeof(int)*n);                int left_max = 0;        int right_max = 0;        int sum = 0;                for(int i = 0;i<n;i++)        {            left_highest[i] = left_max;            left_max = max(left_max,A[i]);        }                for(int i = n-1; i >= 0; i--)        {            right_max = max(right_max,A[i]);            sum += max(0,min(left_highest[i],right_max)-A[i]);        }        return sum;    }};

34 milli secs