acm-1003 求一个数组中连续区间和的最大值问题

本文链接：https://blog.youkuaiyun.com/xiaoxiaosu3/article/details/45176417

本文提供了两种不同的算法实现，一种时间复杂度为O(n)，另一种为O(n^2)。这两种方法用于寻找给定数组中具有最大和的连续子数组，并记录该子数组的起始和结束位置。

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时间复杂度为n的方法：

import java.util.Scanner;

class Main {
    public static void main(String[] args) {
        Scanner sc = new Scanner(System.in);
        int n = sc.nextInt();
        int max = 0, sum1 = 0, first1 = 0, end1 = 0;
        int ar = 0, num = 0;
        for (int i = 0; i < n; i++) {
            num = sc.nextInt();

            for (int j = 0; j < num; j++) {
                ar = sc.nextInt();
                if (sum1 < 0) {
                    sum1 = ar;
                    first1 = j;
                } else {
                    sum1 += ar;
                }
                if (sum1 > max) {
                    max = sum1;
                    end1 = j;
                }

            }
            System.out.println("Case " + (int) (i + 1) + ":");
            System.out.println(max + " " + (int) (first1 + 1) + " "
                    + (int) (end1 + 1));
            max = 0;
            sum1 = 0;
            first1 = 0;
            end1 = 0;
        }

    }
}

时间复杂度为n^2的方法：

import java.util.Scanner;

class Main {
    public static void main(String[] args) {
        Scanner sc = new Scanner(System.in);
        int n = sc.nextInt();
        int sum = 0, sum1 = 0, first = 0, end = 0, first1 = 0, end1 = 0;

        for (int i = 0; i < n; i++) {
            int num = sc.nextInt();
            int[] ar = new int[num];
            for (int j = 0; j < num; j++) {
                ar[j] = sc.nextInt();
            }
            for (int j = 0; j < num; j++) {

                if (ar[j] >= 0) {
                    for (int k = j; k < ar.length; k++) {
                        sum1 += ar[k];
                        if (sum1 < 0)
                            break;
                        if (sum1 > sum) {
                            sum = sum1;
                            first = j;
                            end = k;
                        }
                    }
                    sum1 = 0;
                }

            }
            System.out.println("Case " + (int) (i + 1) + ":");
            System.out.println(sum + " " + (int) (first + 1) + " "
                    + (int) (end + 1));
            sum = 0;
            first = 0;
            end = 0;
        }

    }
}