42. 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!

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分析：

这个题目看得懂别人的解法，却不能证明正确性。

分别维护左右两边的标记l,r，每次去a[l]与a[r]中的小者，假设a[l]<a[r]，对l的左边延伸，若l的右边小于a[l]，则用a[l]去减就是装水的地方。一直延伸到l的右边某个点大于左边，停止。反之亦然。

我觉得需要注意的是，最外层的while判断条件，是等于也没关系，因为里面还会保证相等也不会有影响。

代码：

class Solution {

public:

    int trap(vector<int>& A) {

        int l=0,r=A.size()-1,res=0;

        while(l<r)

    
4000
    {

            int minn=min(A[l],A[r]);

            if(minn==A[l])

            {

                while(++l<r&&A[l]<=minn)

                res+=minn-A[l];

            }

            else

            {

                while(l<--r&&A[r]<=minn)

                res+=minn-A[r];

            }

        }

        return res;

    }

};