1. Two Sum - Easy - Leetcode

Given an array of integers, return indices of the two numbers such that they add up to a specific target.

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

Example:

Given nums = [2, 7, 11, 15], target = 9,

Because nums[0] + nums[1] = 2 + 7 = 9,
return [0, 1].

Related Topic: Array, Hash Table

解题思路：这是LC一道经典题，考虑使用HashMap来保存input中值与其数组index的对应关系。在遍历数组的过程中，若target减去当前遍历值在HashMap中存在，则证明已找到对应的两个值，找到其index返回结果即可，反之则将当前遍历点写入HashMap。

Time Complexity: O(n), Space Complexity: O(n)

Java版本：

 1 class Solution {
 2     public int[] twoSum(int[] nums, int target) {
 3         int[] result = new int[2];
 4         if (nums == null || nums.length < 2) {
 5             return result;
 6         }
 7
 8         Map<Integer, Integer> map = new HashMap<Integer, Integer>();
 9
10         for (int i = 0; i < nums.length; i++) {
11             if (map.containsKey(target - nums[i])) {
12                 result[0] = map.get(target - nums[i]);
13                 result[1] = i;
14             }
15             map.put(nums[i], i);
16         }
17
18         return result;
19     }
20 }

JavaScript版本：

 1 /**
 2  * @param {number[]} nums
 3  * @param {number} target
 4  * @return {number[]}
 5  */
 6 var twoSum = function(nums, target) {
 7     var result = new Array(2);
 8     if (nums == null || nums.length < 2) {
 9         return result;
10     }
11
12     var map = new Map();
13
14     for (var i = 0; i < nums.length; i++) {
15         if (map.has(target - nums[i])) {
16             result[0] = map.get(target - nums[i]);
17             result[1] = i;
18         }
19         map.set(nums[i], i);
20     }
21
22     return result;
23 };

Python版本：

 1 class Solution:
 2     def twoSum(self, nums: List[int], target: int) -> List[int]:
 3         result = [None] * 2
 4         if len(nums) < 2:
 5             return result
 6
 7         dict = {}
 8
 9         for i in range(len(nums)):
10             if (target - nums[i]) in dict:
11                 result[0] = dict.get(target - nums[i])
12                 result[1] = i
13
14             dict[nums[i]] = i
15
16         return result

原文地址：https://www.cnblogs.com/raymondwang/p/10887505.html