[LeetCode] 167. Two Sum II

Given a 1-indexed array of integers numbers that is already sorted in non-decreasing order, find two numbers such that they add up to a specific target number. Let these two numbers be numbers[index1] and numbers[index2] where 1 <= index1 < index2 <= numbers.length.

Return the indices of the two numbers, index1 and index2, added by one as an integer array [index1, index2] of length 2.

The tests are generated such that there is exactly one solution. You may not use the same element twice.

Example 1:

Input: numbers = [2,7,11,15], target = 9
Output: [1,2]
Explanation: The sum of 2 and 7 is 9. Therefore, index1 = 1, index2 = 2. We return [1, 2].

Example 2:

Input: numbers = [2,3,4], target = 6
Output: [1,3]
Explanation: The sum of 2 and 4 is 6. Therefore index1 = 1, index2 = 3. We return [1, 3].

Example 3:

Input: numbers = [-1,0], target = -1
Output: [1,2]
Explanation: The sum of -1 and 0 is -1. Therefore index1 = 1, index2 = 2. We return [1, 2].

Constraints:

• 2 <= numbers.length <= 3 * 104
• -1000 <= numbers[i] <= 1000
• numbers is sorted in non-decreasing order.
• -1000 <= target <= 1000
• The tests are generated such that there is exactly one solution.

两数之和 II - 输入有序数组。

给定一个已按照 非递减顺序排列  的整数数组 numbers ，请你从数组中找出两个数满足相加之和等于目标数 target 。

函数应该以长度为 2 的整数数组的形式返回这两个数的下标值。numbers 的下标 从 1 开始计数 ，所以答案数组应当满足 1 <= answer[0] < answer[1] <= numbers.length 。

你可以假设每个输入 只对应唯一的答案 ，而且你 不可以 重复使用相同的元素。

来源：力扣（LeetCode）
链接：https://leetcode-cn.com/problems/two-sum-ii-input-array-is-sorted
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题意是给一个已经按升序排好序的数组，请返回两个数字的下标，两个数字满足和为 target。因为是已经排好序的 array，所以用 two pointer 从两边往中间逼近，注意最后返回的下标 + 1。

Java实现

class Solution {
    public int[] twoSum(int[] numbers, int target) {
        // corner case
        if (numbers == null || numbers.length < 2) return new int[] { -1, -1};

        // normal case
        int start = 0;
        int end = numbers.length - 1;
        while (start < end) {
            int sum = numbers[start] + numbers[end];
            if (sum == target) {
                return new int[] {start + 1, end + 1};
            } else if (sum < target) {
                start++;
            } else {
                end--;
            }
        }
        return new int[] { -1, -1};
    }
}

JavaScript实现

/**
 * @param {number[]} numbers
 * @param {number} target
 * @return {number[]}
 */
var twoSum = function(numbers, target) {
    // corner case
    if (numbers === undefined || numbers.length < 2) {
        return [-1, -1];
    }

    // normal case
    let start = 0;
    let end = numbers.length - 1;
    while (start < end) {
        const sum = numbers[start] + numbers[end];
        if (sum === target) {
            return [start + 1, end + 1];
        } else if (sum < target) {
            start++;
        } else {
            end--;
        }
    }
    return [-1, -1];
};

LeetCode 题目总结