README_EN.md

42. Trapping Rain Water

中文文档

Description

Given n non-negative integers representing an elevation map where the width of each bar is 1, compute how much water it can trap after raining.

 

Example 1:

Input: height = [0,1,0,2,1,0,1,3,2,1,2,1]
Output: 6
Explanation: The above elevation map (black section) is represented by array [0,1,0,2,1,0,1,3,2,1,2,1]. In this case, 6 units of rain water (blue section) are being trapped.

Example 2:

Input: height = [4,2,0,3,2,5]
Output: 9

 

Constraints:

• n == height.length
• 0 <= n <= 3 * 104
• 0 <= height[i] <= 105

Solutions

Python3

class Solution:
    def trap(self, height: List[int]) -> int:
        n = len(height)
        if n < 3:
            return 0

        left_max = [height[0]] * n
        for i in range(1, n):
            left_max[i] = max(left_max[i - 1], height[i])

        right_max = [height[n - 1]] * n
        for i in range(n - 2, -1, -1):
            right_max[i] = max(right_max[i + 1], height[i])

        res = 0
        for i in range(n):
            res += min(left_max[i], right_max[i]) - height[i]
        return res

Java

class Solution {
    public int trap(int[] height) {
        int n;
        if ((n = height.length) < 3) return 0;

        int[] leftMax = new int[n];
        leftMax[0] = height[0];
        for (int i = 1; i < n; ++i) {
            leftMax[i] = Math.max(leftMax[i - 1], height[i]);
        }

        int[] rightMax = new int[n];
        rightMax[n - 1] = height[n - 1];
        for (int i = n - 2; i >= 0; --i) {
            rightMax[i] = Math.max(rightMax[i + 1], height[i]);
        }

        int res = 0;
        for (int i = 0; i < n; ++i) {
            res += Math.min(leftMax[i], rightMax[i]) - height[i];
        }
        return res;
    }
}

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