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;
}