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.

分析：先找到最高的那个墙height，以及它的下标idx，然后分别往左边和右边，找到次高的墙，这样它们之间围住的水量即为：两堵墙之间的空间减去两堵墙之间存在墙的空间。以此方式进行递归调用，即可就得最终解，AC代码如下：

class Solution {public:    int trap(vector<int>& height) {        if(height.empty())            return 0;        int n = height.size();        int res = 0;        int idx = 0;        int tmp = height[idx];        for(int i = idx + 1; i < n; ++i) {            if(height[i] > tmp) {                tmp = height[i];                idx = i;            }        }        res = leftleave(height, idx, true) + leftleave(height, idx, false);        return res;    }    int leftleave(vector<int>& height, int start, bool isLeft) {        int n = height.size();        // if(start <= 1 || start >= n-2)         //     return 0;        if(isLeft) {            if(start <= 1)                 return 0;            int idx = start-1;            int res = height[idx];            int ans = 0;            for(int i = idx-1; i >= 0; --i) {                if(height[i] > res) {                    res = height[i];                    idx = i;                }            }            for(int i = idx+1; i < start; ++i) {                ans += height[i];            }            return height[idx] * (start - idx - 1) - ans + leftleave(height, idx, true);        }        else {            if(start >= n-2)                return 0;            int idx = start+1;            int res = height[idx];            int ans = 0;            for(int i = idx+1; i < n; ++i) {                if(height[i] > res) {                    res = height[i];                    idx = i;                }            }            for(int i = start+1; i < idx; ++i) {                ans += height[i];            }            return height[idx] * (idx - start - 1) - ans + leftleave(height, idx, false);        }    }};