和为s的数字（Java实现）_java数组中找出和为s的n个数字-CSDN博客

本文介绍了一种高效算法，用于在递增数组中找到和为特定数值S的两个数字，以及打印出所有和为S的连续正数序列。通过双指针技巧，算法在O(n)时间内解决了这两个问题。

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public class E57NumbersWithSum {
    //和为s的数字

    /*问题一：在递增数组中寻找和为s的两个数字*/
    public static void findTwoNumbers(int[] numbers, int s){
        if (numbers == null || numbers.length < 2)
            return;
        int start = 0;
        int end = numbers.length - 1;
        while(start < end){
            long sum = numbers[start] + numbers[end];
            if (sum == s){
                System.out.println(numbers[start] + "," + numbers[end]);
                return;
            }
            else if (sum > s)
                end--;
            else
                start++;
        }
    }

    /*问题二：打印出所有和为s的连续正数序列*/
    public static void findSequence(int s){
        if (s < 3)
            return;
        int pre = 1;
        int behind = 2;
        int middle = (1 + s) >> 1;
        int sum = 3;
        while(pre < middle){
            if (sum == s){
                printSequence(pre, behind);
            }
            while(sum > s && pre < middle){
                sum -= pre;
                pre++;
                if (sum == s){
                    printSequence(pre, behind);
                }
            }
            behind++;
            sum += behind;
        }
    }
    //打印满足要求的连续正数序列
    private static void printSequence(int pre, int behind) {
        if (pre <= 0 || behind <= 0 || pre > behind)
            return;
        System.out.print("[");
        for (int i = pre; i < behind; i++){
            System.out.printf("%d,", i);
        }
        System.out.print(behind + "]\n");
    }

    //测试用例
    public static void main(String[] args){
        int[] numbers1 = {1, 2, 4, 7 ,11, 15};
        E57NumbersWithSum.findTwoNumbers(numbers1, 15); // 4,11
        E57NumbersWithSum.findSequence(15); // 1-5 4-6 7-8
    }
}