标签：row div 一个 ice i++ bec 导入 two sum col
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].
Approch 1: Brute Force 暴风算法
Time complexity : O(n2).
Space complexity : O(1).
class Solution { public int[] twoSum(int[] nums, int target) { for(int i=0; i<nums.length; i++){ for(int j=i+1; j<nums.length;j++){ if((nums[i] + nums[j]) == target) { return new int[]{i,j}; //注意这个return新数组的格式！ } } }
//注意最后没有return的话补加一个异常声明 throw new IllegalArgumentException("No two sum solution"); } }
Approch 2：Two-pass Hash Table
Time complexity : O(n).
Space complexity : O(n).
！！第一次用Map和HashMap（都属于java.util, 实际用的时候需要导入）
标记c为带查找的那个差值，然后用哈希表就能很快get该值对应的在数组中的序号。因此只需要一个for循环就够了。
class Solution { 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 c = target - nums[i]; if(map.containsKey(c)&&map.get(c)!=i){ return new int[]{i,map.get(c)}; } } throw new IllegalArgumentException("No two sum solution"); } }
标签：row div 一个 ice i++ bec 导入 two sum col
原文地址：https://www.cnblogs.com/jamieliu/p/10289594.html
