2024-01-03：用go語言，給你兩個長度為 n 下標從 0 開始的整數數組 cost 和 time，

分別表示給 n 堵不同的墻刷油漆需要的開銷和時間。你有兩名油漆匠，

一位需要 付費 的油漆匠，刷第 i 堵墻需要花費 time[i] 單位的時間，

開銷為 cost[i] 單位的錢。

一位 免費 的油漆匠，刷 任意 一堵墻的時間為 1 單位，開銷為 0，

但是必須在付費油漆匠 工作 時，免費油漆匠才會工作。

請你返回刷完 n 堵墻最少開銷為多少？

輸入：cost = [1,2,3,2], time = [1,2,3,2]。

輸出：3。

來自力扣。2742. 給墻壁刷油漆。

答案2024-01-03：

來自左程云。

靈捷3.5

大體過程如下：

paintWalls1 函數

1.paintWalls1 函數是基于遞歸方法的解決方案。

2.在 process1 函數中，通過遞歸方式將每種情況下的最小開銷計算出來。

3.遞歸調用時考慮兩種情況，選擇當前墻刷或者不刷，計算出最小開銷。

4.該方法在遞歸調用的過程中可能會有很多重復計算，效率可能不高。

paintWalls2 函數

1.paintWalls2 函數采用了記憶化搜索的方式。

2.定義了一個二維數組 dp 用于記錄已經計算過的結果，避免重復計算。

3.通過遞歸+記憶化搜索的方式優化了重復計算，提高了效率。

paintWalls3 函數

1.paintWalls3 函數采用了動態規劃的方式。

2.使用一個一維數組 dp 保存不同墻數下的最小開銷。

3.結合循環和動態遞推的方式，迭代計算每墻的最小開銷，直到第 n 墻。

時間和空間復雜度

• 時間復雜度：

  • paintWalls1 使用了遞歸，可能有大量重復計算，其時間復雜度為 O(2^n)。

  • paintWalls2 和 paintWalls3 使用了記憶化搜索和動態規劃，時間復雜度都為 O(n^2)，其中 n 為墻的數量。

• 空間復雜度：

  • paintWalls1 和 paintWalls2 的額外空間復雜度為 O(n^2)，因為它們都使用了二維數組存儲中間結果。

  • paintWalls3 的額外空間復雜度為 O(n)，因為它只用了一個一維數組保存中間結果。

go完整代碼如下：

package main

import (
    "fmt"
    "math"
)

// paintWalls1 represents the first function from the given Java code.
func paintWalls1(cost []int, time []int) int {
    return process1(cost, time, 0, len(cost))
}

// process1 is the recursive function as mentioned in the Java code.
func process1(cost []int, time []int, i int, s int) int {
    if s <= 0 {
        return 0
    }
    // s > 0
    if i == len(cost) {
        return math.MaxInt32
    } else {
        p1 := process1(cost, time, i+1, s)
        p2 := math.MaxInt32
        next2 := process1(cost, time, i+1, s-1-time[i])
        if next2 != math.MaxInt32 {
            p2 = cost[i] + next2
        }
        return int(math.Min(float64(p1), float64(p2)))
    }
}

// paintWalls2 is the second function from the given Java code.
func paintWalls2(cost []int, time []int) int {
    n := len(cost)
    dp := make([][]int, n+1)
    for i := range dp {
        dp[i] = make([]int, n+1)
        for j := range dp[i] {
            dp[i][j] = -1
        }
    }
    return process2(cost, time, 0, n, dp)
}

// process2 represents the recursive function in the second approach of the Java code.
func process2(cost []int, time []int, i int, s int, dp [][]int) int {
    if s <= 0 {
        return 0
    }
    if dp[i][s] != -1 {
        return dp[i][s]
    }
    var ans int
    if i == len(cost) {
        ans = math.MaxInt32
    } else {
        p1 := process2(cost, time, i+1, s, dp)
        p2 := math.MaxInt32
        next2 := process2(cost, time, i+1, s-1-time[i], dp)
        if next2 != math.MaxInt32 {
            p2 = cost[i] + next2
        }
        ans = int(math.Min(float64(p1), float64(p2)))
    }
    dp[i][s] = ans
    return ans
}

// paintWalls3 is the third function from the given Java code.
func paintWalls3(cost []int, time []int) int {
    n := len(cost)
    dp := make([]int, n+1)
    for i := range dp {
        dp[i] = math.MaxInt32
    }
    dp[0] = 0
    for i := n - 1; i >= 0; i-- {
        for s := n; s >= 1; s-- {
            if s-1-time[i] <= 0 {
                dp[s] = int(math.Min(float64(dp[s]), float64(cost[i])))
            } else if dp[s-1-time[i]] != math.MaxInt32 {
                dp[s] = int(math.Min(float64(dp[s]), float64(cost[i]+dp[s-1-time[i]])))
            }
        }
    }
    return dp[n]
}

func main() {
    cost := []int{1, 2, 3, 2}
    time := []int{1, 2, 3, 2}
    fmt.Println("Result 1:", paintWalls1(cost, time))
    fmt.Println("Result 2:", paintWalls2(cost, time))
    fmt.Println("Result 3:", paintWalls3(cost, time))
}

rust完整代碼如下：

fn paint_walls1(cost: Vec<i32>, time: Vec<i32>) -> i32 {
    process1(&cost, &time, 0, cost.len() as i32)
}

fn process1(cost: &Vec<i32>, time: &Vec<i32>, i: i32, s: i32) -> i32 {
    if s <= 0 {
        return 0;
    }
    if (i as usize) == cost.len() {
        return i32::MAX;
    } else {
        let p1 = process1(cost, time, i + 1, s);
        let mut p2 = i32::MAX;
        let next2 = process1(cost, time, i + 1, s - 1 - time[i as usize]);
        if next2 != i32::MAX {
            p2 = cost[i as usize] + next2;
        }
        return p1.min(p2);
    }
}

fn paint_walls2(cost: Vec<i32>, time: Vec<i32>) -> i32 {
    let n = cost.len();
    let mut dp = vec![vec![-1; n + 1]; n + 1];
    process2(&cost, &time, 0, n as i32, &mut dp)
}

fn process2(cost: &Vec<i32>, time: &Vec<i32>, i: i32, s: i32, dp: &mut Vec<Vec<i32>>) -> i32 {
    if s <= 0 {
        return 0;
    }
    if dp[i as usize][s as usize] != -1 {
        return dp[i as usize][s as usize];
    }
    let ans;
    if (i as usize) == cost.len() {
        ans = i32::MAX;
    } else {
        let p1 = process2(cost, time, i + 1, s, dp);
        let mut p2 = i32::MAX;
        let next2 = process2(cost, time, i + 1, s - 1 - time[i as usize], dp);
        if next2 != i32::MAX {
            p2 = cost[i as usize] + next2;
        }
        ans = p1.min(p2);
    }
    dp[i as usize][s as usize] = ans;
    ans
}

fn paint_walls3(cost: Vec<i32>, time: Vec<i32>) -> i32 {
    let n = cost.len();
    let mut dp = vec![i32::MAX; n + 1];
    dp[0] = 0;
    for i in (0..n).rev() {
        for s in (1..=n as i32).rev() {
            if s - 1 - time[i] <= 0 {
                dp[s as usize] = dp[s as usize].min(cost[i]);
            } else if dp[(s - 1 - time[i]) as usize] != i32::MAX {
                dp[s as usize] = dp[s as usize].min(cost[i] + dp[(s - 1 - time[i]) as usize]);
            }
        }
    }
    dp[n]
}

fn main() {
    let cost = vec![1, 2, 3, 2];
    let time = vec![1, 2, 3, 2];
    
    let result1 = paint_walls1(cost.clone(), time.clone());
    let result2 = paint_walls2(cost.clone(), time.clone());
    let result3 = paint_walls3(cost.clone(), time.clone());

    println!("Result for paint_walls1: {}", result1);
    println!("Result for paint_walls2: {}", result2);
    println!("Result for paint_walls3: {}", result3);
}

c++完整代碼如下：

#include <iostream>
#include <vector>
#include <algorithm>
using namespace std;

// 暴力遞歸
int process1(vector<int>& cost, vector<int>& time, int i, int s) {
    if (s <= 0) {
        return 0;
    }
    if (i == cost.size()) {
        return INT_MAX;
    }
    else {
        int p1 = process1(cost, time, i + 1, s);
        int p2 = INT_MAX;
        int next2 = process1(cost, time, i + 1, s - 1 - time[i]);
        if (next2 != INT_MAX) {
            p2 = cost[i] + next2;
        }
        return min(p1, p2);
    }
}

int paintWalls1(vector<int>& cost, vector<int>& time) {
    return process1(cost, time, 0, cost.size());
}

// 暴力遞歸改記憶化搜索
int process2(vector<int>& cost, vector<int>& time, int i, int s, vector<vector<int>>& dp) {
    if (s <= 0) {
        return 0;
    }
    if (dp[i][s] != -1) {
        return dp[i][s];
    }
    int ans;
    if (i == cost.size()) {
        ans = INT_MAX;
    }
    else {
        int p1 = process2(cost, time, i + 1, s, dp);
        int p2 = INT_MAX;
        int next2 = process2(cost, time, i + 1, s - 1 - time[i], dp);
        if (next2 != INT_MAX) {
            p2 = cost[i] + next2;
        }
        ans = min(p1, p2);
    }
    dp[i][s] = ans;
    return ans;
}

int paintWalls2(vector<int>& cost, vector<int>& time) {
    int n = cost.size();
    vector<vector<int>> dp(n + 1, vector<int>(n + 1, -1));
    return process2(cost, time, 0, n, dp);
}

// 嚴格位置依賴的動態規劃 + 空間壓縮
int paintWalls3(vector<int>& cost, vector<int>& time) {
    int n = cost.size();
    vector<int> dp(n + 1, INT_MAX);
    dp[0] = 0;
    for (int i = n - 1; i >= 0; i--) {
        for (int s = n; s >= 1; s--) {
            if (s - 1 - time[i] <= 0) {
                dp[s] = min(dp[s], cost[i]);
            }
            else if (dp[s - 1 - time[i]] != INT_MAX) {
                dp[s] = min(dp[s], cost[i] + dp[s - 1 - time[i]]);
            }
        }
    }
    return dp[n];
}

int main() {
    vector<int> cost = { 1, 2, 3, 2 };
    vector<int> time = { 1, 2, 3, 2 };

    cout << "Result for paintWalls1: " << paintWalls1(cost, time) << endl;
    cout << "Result for paintWalls2: " << paintWalls2(cost, time) << endl;
    cout << "Result for paintWalls3: " << paintWalls3(cost, time) << endl;

    return 0;
}

c語言完整代碼如下：

#include <stdio.h>
#include <stdlib.h>
#include <limits.h>

int process1(int* cost, int* time, int i, int s, int costSize);

int paintWalls1(int* cost, int costSize, int* time, int timeSize) {
    return process1(cost, time, 0, costSize, costSize);
}

int process1(int* cost, int* time, int i, int s, int costSize) {
    if (s <= 0) {
        return 0;
    }
    if (i == costSize) {
        return INT_MAX;
    }
    else {
        int p1 = process1(cost, time, i + 1, s, costSize);
        int p2 = INT_MAX;
        int next2 = process1(cost, time, i + 1, s - 1 - time[i], costSize);
        if (next2 != INT_MAX) {
            p2 = cost[i] + next2;
        }
        return (p1 < p2) ? p1 : p2;
    }
}

int process2(int* cost, int* time, int i, int s, int costSize, int** dp);

int paintWalls2(int* cost, int costSize, int* time, int timeSize) {
    int** dp = (int**)malloc((costSize + 1) * sizeof(int*));
    for (int i = 0; i <= costSize; i++) {
        dp[i] = (int*)malloc((costSize + 1) * sizeof(int));
        for (int j = 0; j <= costSize; j++) {
            dp[i][j] = -1;
        }
    }
    int result = process2(cost, time, 0, costSize, costSize, dp);
    for (int i = 0; i <= costSize; i++) {
        free(dp[i]);
    }
    free(dp);
    return result;
}

int process2(int* cost, int* time, int i, int s, int costSize, int** dp) {
    if (s <= 0) {
        return 0;
    }
    if (dp[i][s] != -1) {
        return dp[i][s];
    }
    int ans;
    if (i == costSize) {
        ans = INT_MAX;
    }
    else {
        int p1 = process2(cost, time, i + 1, s, costSize, dp);
        int p2 = INT_MAX;
        int next2 = process2(cost, time, i + 1, s - 1 - time[i], costSize, dp);
        if (next2 != INT_MAX) {
            p2 = cost[i] + next2;
        }
        ans = (p1 < p2) ? p1 : p2;
    }
    dp[i][s] = ans;
    return ans;
}

int paintWalls3(int* cost, int costSize, int* time, int timeSize);

int paintWalls3(int* cost, int costSize, int* time, int timeSize) {
    int* dp = (int*)malloc((costSize + 1) * sizeof(int));
    for (int i = 0; i <= costSize; i++) {
        dp[i] = INT_MAX;
    }
    dp[0] = 0;
    for (int i = costSize - 1; i >= 0; i--) {
        for (int s = costSize; s >= 1; s--) {
            if (s - 1 - time[i] <= 0) {
                dp[s] = (dp[s] < cost[i]) ? dp[s] : cost[i];
            }
            else if (dp[s - 1 - time[i]] != INT_MAX) {
                dp[s] = (dp[s] < cost[i] + dp[s - 1 - time[i]]) ? dp[s] : cost[i] + dp[s - 1 - time[i]];
            }
        }
    }
    int result = dp[costSize];
    free(dp);
    return result;
}

int main() {
    int cost[] = { 1, 2, 3, 2 };
    int time[] = { 1, 2, 3, 2 };

    int result1 = paintWalls1(cost, 4, time, 4);
    printf("Result of paintWalls1: %d\n", result1);

    int result2 = paintWalls2(cost, 4, time, 4);
    printf("Result of paintWalls2: %d\n", result2);

    int result3 = paintWalls3(cost, 4, time, 4);
    printf("Result of paintWalls3: %d\n", result3);
    return 0;
}

總結

以上是生活随笔為你收集整理的2024-01-03：用go语言，给你两个长度为 n 下标从 0 开始的整数数组 cost 和 time， 分别表示给 n 堵不同的墙刷油漆需要的开销和时间。你有两名油漆匠， 一位需要 付费 的油漆匠的全部內容，希望文章能夠幫你解決所遇到的問題。

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