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], return6.

class Solution {
public:
    int trap(int A[], int n) {
        int secHight = 0;
        int left = 0;
        int right = n-1;
        int area = 0;
        while (left < right){
            if (A[left] < A[right]){
                secHight = max(A[left], secHight);
                area += secHight-A[left];//计算当前格的能装雨水的容量
                left++;
            } else {
                secHight = max(A[right], secHight);
                area += secHight-A[right];
                right--;
            }
        }
        return area;
    }
};

/*同样这种思路，只是分开写了

int trap(int A[], int n) {
        //找到最高的对应的下标
        int highestIndex=0;
        for(int i=0;i<n;i++)
        {
            if(A[highestIndex]<A[i])
                highestIndex=i;
        }
        //计算最高的左边可容纳的水
        int left=0;
        int sumWater=0;
        for(int i=0;i<highestIndex;i++)
        {
            if(left<A[i])
                left=A[i];
            else
            {
                sumWater+=(left-A[i]);
            }
        }
        //计算最高的右边可容纳的水
        int right=0;
        for(int i=n-1;i>highestIndex;i--)
        {
            if(right<A[i])
                right=A[i];
            else
            {
                sumWater+=(right-A[i]);
            }
        }
        return sumWater;
    }

    */