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; }