[LintCode] Trapping Rain Water 收集雨水

 

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.

Have you met this question in a real interview? 

Yes

Example

Given [0,1,0,2,1,0,1,3,2,1,2,1], return 6.

Challenge 

O(n) time and O(1) memory

O(n) time and O(n) memory is also acceptable.

 

LeetCode上的原题，请参见我之前的博客Trapping Rain Water。

 

解法一：

class Solution {
public:
    /**
     * @param heights: a vector of integers
     * @return: a integer
     */
    int trapRainWater(vector<int> &heights) {
        int res = 0, mx = 0, n = heights.size();
        vector<int> dp(n, 0);
        for (int i = 0; i < n; ++i) {
            dp[i] = mx;
            mx = max(mx, heights[i]);
        }
        mx = 0;
        for (int i = n - 1; i >= 0; --i) {
            dp[i] = min(dp[i], mx);
            mx = max(mx, heights[i]);
            if (dp[i] > heights[i]) res += dp[i] - heights[i];
        }
        return res;
    }
};

 

解法二：

class Solution {
public:
    /**
     * @param heights: a vector of integers
     * @return: a integer
     */
    int trapRainWater(vector<int> &heights) {
        int res = 0, l = 0, r = heights.size() - 1;
        while (l < r) {
            int mn = min(heights[l], heights[r]);
            if (mn == heights[l]) {
                ++l;
                while (l < r && heights[l] < mn) {
                    res += mn - heights[l++];
                }
            } else {
                --r;
                while (l < r && heights[r] < mn) {
                    res += mn - heights[r--];
                }
            }
        }
        return res;
    }
};

 

解法三：

class Solution {
public:
    /**
     * @param heights: a vector of integers
     * @return: a integer
     */
    int trapRainWater(vector<int> &heights) {
        int res = 0, l = 0, r = heights.size() - 1, level = 0;
        while (l < r) {
            int lower = heights[(heights[l] < heights[r]) ? l++ : r--];
            level = max(level, lower);
            res += level - lower;
        }
        return res;
    }
};