[Leetcode] trapping rain water 收集雨水

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.

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!

题意：求所有能收集到的雨水

思路：

代码如下：

class Solution {
public:
    int trap(int A[], int n) 
    {
        int maxIdx=0;
        for(int i=0;i<n;++i)
        {
            if(A[i]>A[maxIdx])
                maxIdx=i;
        }

        int left=0,right=0;
        int sum=0;
        for(int i=0;i<maxIdx;++i)   //计算左边容量
        {
            if(left>A[i])
                sum+=left-A[i];
            else
                left=A[i];
        }
        for(int i=n-1;i>maxIdx;i--)     //加上右边容量
        {
            if(right>A[i])
                sum+=right-A[i];
            else
                right=A[i];
        }

        return sum;    
    }
};

方法二：这题和最大水容器有点类似。定义两个指针，分别指向数组的两端，然后从两边向中间变历。具体思路：取两端中值较小（lower）的那段开始向中间遍历，若下一比lower小，则sum加上两者差值就行，left++，对应的值再和lower比较，若还是小于则继续加上差值，若不小于了，则将该值重新和right对应的值比较，重新确定lower值。重复上述过程。这里要注意的是：若连续是左端（右端）为较小值那端，Lower应该是固定的，直到条件不满足。代码如下：

class Solution {
public:
    int trap(int A[], int n) 
    {
        int left=0,right=n-1;
        int sum=0;

        while(left<right)
        {
            int lower=min(left,right);
            if(A[left]<A[right])
            {
                left++;
                while(left<right&&A[left]<lower)
                    sum+=A[lower]-A[left];
            }
            else
            {
                right--;
                while(left<right&&A[right]<lower)
                    sum+=A[lower]-A[right];
            }
        }
        return sum;
    }
};