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!

很有意思的一道题，借鉴大神的思路：用两个数组分别记录某点左边最大高度，右边最大高度，然后取最小值，最后求出最小值与该点的实际高度只差，若大于0则原来计算的储水量增加该差值，否则，不增加。

int trap(int* height, int heightSize) {    if(height==NULL || heightSize<1)return 0;      int leftMax[1000];    int rightMax[1000];    int max = 0;    int water=0;    int i;    for(i = 0;i<heightSize;i++){        leftMax[i] = max;        max = max>height[i]?max:height[i];    }    max = 0;    for(int j =heightSize -1;j>=0;j--){        rightMax[j] = max;        max = max>height[j]?max:height[j];    }    for(i=0;i<heightSize;i++){        int min = leftMax[i]>rightMax[i]?rightMax[i]:leftMax[i];        int h = min - height[i];        if(h>0)           water+=h;    }    return water;}