comments	difficulty	edit_url	tags
true	困难	https://github.com/doocs/leetcode/edit/main/solution/0000-0099/0042.Trapping%20Rain%20Water/README.md	栈 数组 双指针 动态规划 单调栈

42. 接雨水

English Version

题目描述

给定 n 个非负整数表示每个宽度为 1 的柱子的高度图，计算按此排列的柱子，下雨之后能接多少雨水。

 

示例 1：

输入：height = [0,1,0,2,1,0,1,3,2,1,2,1]
输出：6
解释：上面是由数组 [0,1,0,2,1,0,1,3,2,1,2,1] 表示的高度图，在这种情况下，可以接 6 个单位的雨水（蓝色部分表示雨水）。 

示例 2：

输入：height = [4,2,0,3,2,5]
输出：9

 

提示：

• n == height.length
• 1 <= n <= 2 * 104
• 0 <= height[i] <= 105

解法

方法一：动态规划

我们定义 $left[i]$ 表示下标 $i$ 位置及其左边的最高柱子的高度，定义 $right[i]$ 表示下标 $i$ 位置及其右边的最高柱子的高度。那么下标 $i$ 位置能接的雨水量为 $min(left[i],right[i])-height[i]$。我们遍历数组，计算出 $left[i]$ 和 $right[i]$，最后答案为 $\sum _{i=0}^{n-1}min(left[i],right[i])-height[i]$。

时间复杂度 $O(n)$，空间复杂度 $O(n)$。其中 $n$ 为数组的长度。

Python3

class Solution:
    def trap(self, height: List[int]) -> int:
        n = len(height)
        left = [height[0]] * n
        right = [height[-1]] * n
        for i in range(1, n):
            left[i] = max(left[i - 1], height[i])
            right[n - i - 1] = max(right[n - i], height[n - i - 1])
        return sum(min(l, r) - h for l, r, h in zip(left, right, height))

Java

class Solution {
    public int trap(int[] height) {
        int n = height.length;
        int[] left = new int[n];
        int[] right = new int[n];
        left[0] = height[0];
        right[n - 1] = height[n - 1];
        for (int i = 1; i < n; ++i) {
            left[i] = Math.max(left[i - 1], height[i]);
            right[n - i - 1] = Math.max(right[n - i], height[n - i - 1]);
        }
        int ans = 0;
        for (int i = 0; i < n; ++i) {
            ans += Math.min(left[i], right[i]) - height[i];
        }
        return ans;
    }
}

C++

class Solution {
public:
    int trap(vector<int>& height) {
        int n = height.size();
        int left[n], right[n];
        left[0] = height[0];
        right[n - 1] = height[n - 1];
        for (int i = 1; i < n; ++i) {
            left[i] = max(left[i - 1], height[i]);
            right[n - i - 1] = max(right[n - i], height[n - i - 1]);
        }
        int ans = 0;
        for (int i = 0; i < n; ++i) {
            ans += min(left[i], right[i]) - height[i];
        }
        return ans;
    }
};

Go

func trap(height []int) (ans int) {
	n := len(height)
	left := make([]int, n)
	right := make([]int, n)
	left[0], right[n-1] = height[0], height[n-1]
	for i := 1; i < n; i++ {
		left[i] = max(left[i-1], height[i])
		right[n-i-1] = max(right[n-i], height[n-i-1])
	}
	for i, h := range height {
		ans += min(left[i], right[i]) - h
	}
	return
}

TypeScript

function trap(height: number[]): number {
    const n = height.length;
    const left: number[] = new Array(n).fill(height[0]);
    const right: number[] = new Array(n).fill(height[n - 1]);
    for (let i = 1; i < n; ++i) {
        left[i] = Math.max(left[i - 1], height[i]);
        right[n - i - 1] = Math.max(right[n - i], height[n - i - 1]);
    }
    let ans = 0;
    for (let i = 0; i < n; ++i) {
        ans += Math.min(left[i], right[i]) - height[i];
    }
    return ans;
}

Rust

impl Solution {
    #[allow(dead_code)]
    pub fn trap(height: Vec<i32>) -> i32 {
        let n = height.len();
        let mut left: Vec<i32> = vec![0; n];
        let mut right: Vec<i32> = vec![0; n];

        left[0] = height[0];
        right[n - 1] = height[n - 1];

        // Initialize the left & right vector
        for i in 1..n {
            left[i] = std::cmp::max(left[i - 1], height[i]);
            right[n - i - 1] = std::cmp::max(right[n - i], height[n - i - 1]);
        }

        let mut ans = 0;

        // Calculate the ans
        for i in 0..n {
            ans += std::cmp::min(left[i], right[i]) - height[i];
        }

        ans
    }
}

C#

public class Solution {
    public int Trap(int[] height) {
        int n = height.Length;
        int[] left = new int[n];
        int[] right = new int[n];
        left[0] = height[0];
        right[n - 1] = height[n - 1];
        for (int i = 1; i < n; ++i) {
            left[i] = Math.Max(left[i - 1], height[i]);
            right[n - i - 1] = Math.Max(right[n - i], height[n - i - 1]);
        }
        int ans = 0;
        for (int i = 0; i < n; ++i) {
            ans += Math.Min(left[i], right[i]) - height[i];
        }
        return ans;
    }
}

PHP

class Solution {
    /**
     * @param integer[] $height
     * @return integer
     */

    function trap($height) {
        $n = count($height);

        if ($n == 0) {
            return 0;
        }

        $left = 0;
        $right = $n - 1;
        $leftMax = 0;
        $rightMax = 0;
        $ans = 0;

        while ($left < $right) {
            if ($height[$left] < $height[$right]) {
                if ($height[$left] > $leftMax) {
                    $leftMax = $height[$left];
                } else {
                    $ans += $leftMax - $height[$left];
                }
                $left++;
            } else {
                if ($height[$right] > $rightMax) {
                    $rightMax = $height[$right];
                } else {
                    $ans += $rightMax - $height[$right];
                }
                $right--;
            }
        }
        return $ans;
    }
}