Question

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。

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))

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;
  }
}

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;
  }
};

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
}

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;
}

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
  }
}

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;
  }
}