leetcode167. Two Sum II - Input array is sorted

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最新推荐文章于 2025-08-11 15:06:51 发布

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分类专栏： leetcode 文章标签： leetcode TwoPoint

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本文介绍了一种在已排序的整数数组中寻找两个数使其和等于特定目标值的方法。利用双指针技巧，从数组两端开始向中间逼近，高效地找到符合条件的元素对。该算法时间复杂度为O(n)，空间复杂度为O(1)。

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167. Two Sum II - Input array is sorted

Given an array of integers that is already sorted in ascending order, find two numbers such that they add up to a specific target number.

The function twoSum should return indices of the two numbers such that they add up to the target, where index1 must be less than index2. Please note that your returned answers (both index1 and index2) are not zero-based.

You may assume that each input would have exactly one solution and you may not use the same element twice.

Input: numbers={2, 7, 11, 15}, target=9
Output: index1=1, index2=2

解法一

双指针法

public class Solution {
    public int[] twoSum(int[] numbers, int target) {
        int[] ret = new int[2];
        if (numbers == null || numbers.length == 0) {
            return ret;
        }

        int i = 0;
        int j = numbers.length - 1;
        while (i < j) {
            if (numbers[i] + numbers[j] == target) {
                ret[0] = i + 1;
                ret[1] = j + 1;
                break;
            } else if (numbers[i] + numbers[j] > target) {
                j--;
            } else {
                i++;
            }
        }

        return ret;
    }
}

这里写图片描述

解法二

class Solution {
    /**
     * @time O(n)
     * @space O(1)
     */
    public int[] twoSum(int[] numbers, int target) {
        if (numbers == null || numbers.length == 0) {
            return null;
        }

        int i = 0, j = numbers.length - 1;
        while (i < j) {
            if (numbers[i] + numbers[j] == target) {
                return new int[] {i + 1, j + 1};
            } else if (numbers[i] + numbers[j] < target) {
                i++;
            } else {
                j--;
            }
        }

        return null;
    }
}