[Swift]LeetCode167. 两数之和 II

★★★★★★★★★★★★★★★★★★★★★★★★★★★★★★★★★★★★★★★★
➤微信公众号：山青咏芝（shanqingyongzhi）
➤博客园地址：山青咏芝（https://www.cnblogs.com/strengthen/）
➤GitHub地址：https://github.com/strengthen/LeetCode
➤原文地址：https://www.cnblogs.com/strengthen/p/9715077.html 
➤如果链接不是山青咏芝的博客园地址，则可能是爬取作者的文章。
➤原文已修改更新！强烈建议点击原文地址阅读！支持作者！支持原创！
★★★★★★★★★★★★★★★★★★★★★★★★★★★★★★★★★★★★★★★★

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.

Note:

• 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.

Example:

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.

---

给定一个已按照升序排列 的有序数组，找到两个数使得它们相加之和等于目标数。

函数应该返回这两个下标值 index1 和 index2，其中 index1 必须小于 index2。

说明:

• 返回的下标值（index1 和 index2）不是从零开始的。
• 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。

示例:

输入: numbers = [2, 7, 11, 15], target = 9
输出: [1,2]
解释: 2 与 7 之和等于目标数 9 。因此 index1 = 1, index2 = 2 。

---

class Solution {
    func twoSum(_ numbers: [Int], _ target: Int) -> [Int] {
        var leftIndex:Int = 0
        var rightIndex:Int = numbers.count - 1
        while (leftIndex < rightIndex)
        {
            var sum = numbers[leftIndex] + numbers[rightIndex]
            if sum == target 
            {
                return [leftIndex + 1,rightIndex + 1]
            }
            else if sum < target
            {
                leftIndex += 1
            }
            else
            {
                rightIndex -= 1
            }
        }
        return []
    }
}

---

16ms

class Solution {
    func twoSum(_ numbers: [Int], _ target: Int) -> [Int] {
        guard numbers.count >= 2 else {
            return []
        }
        var dic: [Int:Int] = [:]
        for i in 0..<numbers.count {
            let t = target - numbers[i]
            if dic.keys.contains(t) {
                return [dic[t]!+1, i+1]
            }
            dic[numbers[i]] = i
        }
        return []
    }
}

---

24ms

class Solution {
    func twoSum(_ numbers: [Int], _ target: Int) -> [Int] {
        var numIndexDict = [Int : Int]()
        for (index, num) in numbers.enumerated() {
            numIndexDict[num] = index
        }
        
        for (index, num) in numbers.enumerated() {
            if let diffIndex = numIndexDict[target - num] {
                return [index + 1, diffIndex + 1]
            }
        }
        
        return [Int]()
    }
}