标签：because target leetcode [1] contains 思路 length spec ash
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
标签：because target leetcode [1] contains 思路 length spec ash
原文地址：https://www.cnblogs.com/raymondwang/p/10887505.html
