LeetCode——Trapping Rain Water

• 題目
  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)。對於區間[l, r]，如果max(height[l+1] … height[r-1]) > max(height[l], height[r])，那麼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)複雜度。考慮一下，填滿後的特點，一定是中間高兩邊低（邊界情況是遞增或者遞減）。那麼從兩側往中間掃描的時候，就可以知道當前位置的水位。

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;
    }
};