Trapping Rain Water

题目如下：

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.

解析思路：先找到最高峰，然后从两边向中间计算，因为两边向中间是递增的趋势

参考代码如下：

    int trap(int A[], int n) {if(n <= 2) return 0;int max = -1, maxInd = 0;int i = 0;for(; i < n; ++i){//先找出最高峰if(A[i] > max){max = A[i];maxInd = i;}}int area = 0, root = A[0];for(i = 0; i < maxInd; ++i){//从左侧到最高峰的计算if(root < A[i]) root = A[i];else area += (root - A[i]);}for(i = n-1, root = A[n-1]; i > maxInd; --i){//从右侧到最高峰的计算if(root < A[i]) root = A[i];else area += (root - A[i]);}return area;}

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