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 same element twice.
Example:
Given nums = [2, 7, 11, 15], target = 9, Because nums[0] + nums[1] = 2 + 7 = 9, return [0, 1].
public int[] twoSum(int[] nums, int target) {//my int[] re=new int[2]; int sign=0;//找到的标记(结束标记) for(int i=0;i<nums.length-1;i++){ if (1==sign){ break; } for(int j=i+1;j< nums.length;j++){ if(nums[i]+nums[j]==target){ re[0]=i; re[1]=j; sign=1; break; } } } return re; }
时间复杂度为O(n)的方法
public int[] twoSum(int[] nums, int target) { Map<Integer, Integer> map = new HashMap<>(); for (int i = 0; i < nums.length; i++) { int complement = target - nums[i]; if (map.containsKey(complement)) { return new int[] { map.get(complement), i }; } map.put(nums[i], i); } throw new IllegalArgumentException("No two sum solution"); }