leetcode 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

.



每个点是否被水漫过，取决于其前面是否存在高度大于它当前高度的点并且其后也存在高度大于当前高度的点，因此我们可以首先预处理处1->i点中高度最大的点和i->n点中高度最大的点，然后用于最终的ans计算。

class Solution {
public:
    int trap(vector<int>& height) {
        int ans = 0;
        int size = height.size();
        if(size <= 1) return 0;
        vector<int> v1(size), v2(size);
        v1[0] = height[0];
        for(int i = 1; i < size; ++i) {
            v1[i] = max(v1[i-1], height[i]);
        }
        v2[size-1] = height[size-1];
        for(int i = size - 2; i >= 0; --i) {
            v2[i] = max(v2[i+1], height[i]);
        }
        for(int i = 0; i < size; ++i) {
            int h = min(v1[i], v2[i]);
            if(height[i] < h) {
                ans += h - height[i];
            }
        }
        return ans;
    }
};