LeetCode——Trapping Rain Water

• 题目
  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.

• 解法1
  分治法，O(n log n)。对于区间[l, r]，如果max(height[l+1] … height[r-1]) > max(height[l], height[r])，那么res(l, r) = res(l, heighest_index) + res(heighest_index, r)。

class Solution {    int f(vector<int>& height, int l, int r) {        if (r - l <= 1)            return 0;        int Max = -1, id = 0;        for (int i = l + 1; i < r; i++) {            if (height[i] > Max) {                Max = height[i];                id = i;            }        }        if (Max <= height[l] && Max <= height[r]) {            int h = std::min(height[l], height[r]), ret = 0;            for (int i = l + 1; i < r; i++)                ret += std::max(0, h - height[i]);            return ret;        }        else {            return f(height, l, id) + f(height, id, r);        }    }public:    int trap(vector<int>& height) {        int l = 0, r = (int)height.size() - 1;        return f(height, l, r);    }};

• 解法2
  O(n)复杂度。考虑一下，填满后的特点，一定是中间高两边低（边界情况是递增或者递减）。那么从两侧往中间扫描的时候，就可以知道当前位置的水位。

class Solution {public:    int trap(vector<int>& height) {        int l = 0, r = (int)height.size() - 1, level = 0, ret = 0;        while (l < r) {            int lower = height[height[l] < height[r] ? l++ : r--];            level = std::max(level, lower);            ret += level - lower;        }        return ret;    }};