(B卷,100分)- 最长连续子序列（Java & JS & Python & C）

题目描述

有N个正整数组成的一个序列。给定整数sum，求长度最长的连续子序列，使他们的和等于sum，返回此子序列的长度，

如果没有满足要求的序列，返回-1。

输入描述

第一行输入是：N个正整数组成的一个序列

第二行输入是：给定整数sum

输出描述

最长的连续子序列的长度

备注

• 输入序列仅由数字和英文逗号构成，数字之间采用英文逗号分隔
• 序列长度：1 <= N <= 200
• 输入序列不考虑异常情况

用例

前缀和解法

本题数量级不大，可以考虑求解输入序列的前缀和数组，实现O(1)时间复杂度计算任意区间和，关于前缀和可以看：

JS算法源码

const rl = require("readline").createInterface({ input: process.stdin });
var iter = rl[Symbol.asyncIterator]();
const readline = async () => (await iter.next()).value;

void (async function () {
  const nums = (await readline()).split(",").map(Number);
  const target = parseInt(await readline());

  console.log(getResult(nums, target));
})();

function getResult(nums, target) {
  const n = nums.length;

  const preSum = new Array(n + 1).fill(0);
  for (let i = 1; i <= n; i++) {
    preSum[i] = preSum[i - 1] + nums[i - 1];
  }

  let maxLen = -1;
  for (let l = 0; l <= n - 1; l++) {
    for (let r = l + 1; r <= n; r++) {
      if (preSum[r] - preSum[l] == target) {
        maxLen = Math.max(maxLen, r - l);
      }
    }
  }

  return maxLen;
}

Java算法源码

import java.util.Arrays;
import java.util.Scanner;

public class Main {
  public static void main(String[] args) {
    Scanner sc = new Scanner(System.in);

    int[] nums = Arrays.stream(sc.nextLine().split(",")).mapToInt(Integer::parseInt).toArray();
    int target = Integer.parseInt(sc.nextLine());

    System.out.println(getResult(nums, target));
  }

  public static int getResult(int[] nums, int target) {
    int n = nums.length;

    // 前缀和数组
    int[] preSum = new int[n + 1];
    for (int i = 1; i <= n; i++) {
      preSum[i] = preSum[i - 1] + nums[i - 1];
    }

    // 记录最长连续子序列的长度
    int maxLen = -1;

    // 求解所有连续子序列的区间和
    for (int l = 0; l <= n - 1; l++) {
      for (int r = l + 1; r <= n; r++) {
        if (preSum[r] - preSum[l] == target) {
          maxLen = Math.max(maxLen, r - l);
        }
      }
    }

    return maxLen;
  }
}

Python算法源码

# 输入获取
nums = list(map(int, input().split(",")))
target = int(input())


# 算法入口
def getResult():
    n = len(nums)

    preSum = [0] * (n + 1)
    for i in range(1, n + 1):
        preSum[i] = preSum[i - 1] + nums[i - 1]

    maxLen = -1

    for l in range(0, n):
        for r in range(l + 1, n+1):
            if preSum[r] - preSum[l] == target:
                maxLen = max(maxLen, r - l)

    return maxLen


# 算法调用
print(getResult())

C算法源码

#include <stdio.h>

#define MAX(a,b) (a) > (b) ? (a) : (b)

#define MAX_SIZE 200

int getResult(int nums[], int nums_size, int target);

int main() {
    int nums[MAX_SIZE];
    int nums_size = 0;

    while(scanf("%d", &nums[nums_size++])) {
        if(getchar() != ',') break;
    }

    int target;
    scanf("%d", &target);

    printf("%dn", getResult(nums, nums_size, target));

    return 0;
}

int getResult(int nums[], int nums_size, int target) {
    int preSum[nums_size+1];

    preSum[0] = 0;
    for(int i=1; i<=nums_size; i++) {
        preSum[i] = preSum[i-1] + nums[i-1];
    }

    int maxLen = -1;
    for(int l=0; l<= nums_size-1; l++) {
        for(int r=l+1; r <= nums_size; r++) {
            if(preSum[r] - preSum[l] == target) {
                maxLen = MAX(maxLen, r - l);
            }
        }
    }

    return maxLen;
}

双指针解法

本题可以利用双指针解决。

 同时，有一种特殊情况

2023.06.12

本题的第二行输入是：给定整数sum

因此sum可能取负数，0，正数，对于sum为负数，0的情况，可以直接返回-1，因为第一行输入是：N个正整数组成的一个序列

JavaScript算法源码

/* JavaScript Node ACM模式 控制台输入获取 */
const readline = require("readline");

const rl = readline.createInterface({
  input: process.stdin,
  output: process.stdout,
});

const lines = [];
rl.on("line", (line) => {
  lines.push(line);

  if (lines.length === 2) {
    const arr = lines[0].split(",").map(Number);
    const sum = parseInt(lines[1]);

    console.log(getMaxLen(arr, sum));

    lines.length = 0;
  }
});

function getMaxLen(arr, sum) {
  let ans = -1;

  if (sum <= 0) return ans;

  let l = 0;
  let r = 0;
  let n = arr.length;

  let total = arr[l];

  while (true) {
    if (total > sum) {
      l++;
      total -= arr[l - 1];
      if (l < n && r < l) {
        r = l;
        total += arr[r];
      }
    } else if (total < sum) {
      r++;
      if (r < n) total += arr[r];
      else break;
    } else {
      ans = Math.max(ans, r - l + 1);
      l++;
      r++;
      if (r < n) total += arr[r] - arr[l - 1];
      else break;
    }
  }

  return ans;
}

Java算法源码

import java.util.Arrays;
import java.util.Scanner;

public class Main {
  public static void main(String[] args) {
    Scanner sc = new Scanner(System.in);

    Integer[] arr =
        Arrays.stream(sc.nextLine().split(",")).map(Integer::parseInt).toArray(Integer[]::new);

    int sum = Integer.parseInt(sc.nextLine());

    System.out.println(getResult(arr, sum));
  }

  public static int getResult(Integer[] arr, int sum) {
    int ans = -1;
    if (sum <= 0) return ans;

    int l = 0;
    int r = 0;
    int n = arr.length;

    int total = arr[l];
    while (true) {
      if (total > sum) {
        l++;
        total -= arr[l - 1];
        if (l < n && r < l) {
          r = l;
          total += arr[r];
        }
      } else if (total < sum) {
        r++;
        if (r < n) total += arr[r];
        else break;
      } else {
        ans = Math.max(ans, r - l + 1);
        l++;
        r++;
        if (r < n) total += arr[r] - arr[l - 1];
        else break;
      }
    }

    return ans;
  }
}

Python算法源码

# 输入获取
arr = list(map(int, input().split(",")))
sumV = int(input())


# 算法入口
def getResult():
    ans = -1

    if sumV <= 0:
        return ans

    l = 0
    r = 0
    n = len(arr)

    total = arr[l]

    while True:
        if total > sumV:
            l += 1
            total -= arr[l - 1]
            if r < l < n:
                r = l
                total += arr[r]
        elif total < sumV:
            r += 1
            if r < n:
                total += arr[r]
            else:
                break
        else:
            ans = max(ans, r - l + 1)
            l += 1
            r += 1
            if r < n:
                total += arr[r] - arr[l - 1]
            else:
                break

    return ans


# 算法调用
print(getResult())

C算法源码

#include <stdio.h>

#define MAX(a, b) (a) > (b) ? (a) : (b)

#define MAX_SIZE 200

int getResult(int nums[], int nums_size, int target);

int main() {
    int nums[MAX_SIZE];
    int nums_size = 0;

    while (scanf("%d", &nums[nums_size++])) {
        if (getchar() != ',') break;
    }

    int target;
    scanf("%d", &target);

    printf("%dn", getResult(nums, nums_size, target));

    return 0;
}

int getResult(int nums[], int nums_size, int target) {
    int maxLen = -1;
    if (target <= 0) return maxLen;

    int l = 0;
    int r = 0;

    int sum = nums[l];
    while (1) {
        if (sum > target) {
            l++;
            sum -= nums[l - 1];

            if (l < nums_size && r < l) {
                r = l;
                sum += nums[r];
            }
        } else if (sum < target) {
            r++;
            if (r < nums_size) sum += nums[r];
            else break;
        } else {
            maxLen = MAX(maxLen, r - l + 1);
            l++;
            r++;
            if (r < nums_size) sum += nums[r] - nums[l - 1];
            else break;
        }
    }

    return maxLen;
}