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. Example:
Given nums = [2, 7, 11, 15], target = 9, Because nums[0] + nums[1] = 2 + 7 = 9, return [0, 1].题目意思是给定一个数组以及目标值，找出数字中对应的两个元素的索引。 方法1： 我的方法—>暴力方法，循环比较所有两两元素和的比较。
public class Solution { public int[] twoSum(int[] nums, int target) { int size=nums.length; int [] num=new int[2]; int sum; for(int i=0;i<size-1;i++) for(int j=i+1;j<size;j++) { if(nums[i]+nums[j]==target) { num[0]=i; num[1]=j; } } return num; } }时间复杂度O(n2)，空间复杂度O(1)
方法2： 哈希表法：
public int[] twoSum(int[] nums, int target) { Map<Integer, Integer> map = new HashMap<>(); for (int i = 0; i < nums.length; i++) { map.put(nums[i], i); } for (int i = 0; i < nums.length; i++) { int complement = target - nums[i]; if (map.containsKey(complement) && map.get(complement) != i) { return new int[] { i, map.get(complement) }; } } throw new IllegalArgumentException("No two sum solution"); }时间复杂度O(n)
