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)
        {
            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;
    }
};