LeetCode167. Two Sum II - Input array is sorted

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本文介绍了一种针对已排序数组查找两个数使其和为目标值的高效算法。该算法采用两端逼近的方式，时间复杂度为O(n)，空间复杂度为O(1)。

题目

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

最初想法

两个for循环，O(n²)时间复杂度；
HashMap，先把所有数放进去，然后从头找 target - numbers[i]，O(n)时间复杂度，但是开销大，也很慢。

O(n)时间复杂度，O(1)时间复杂度方法，两头找

直接上代码

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