标签：row inpu public shm except and contain turn new
Description:
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].
My Solution：
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}; } } } throw new IllegalArgumentException("no two sum solution!"); } }
Better Solution:
class Solution { //用hashMap存储{target - nums[i]:i},如果包含target-nums[i]这个key，说明找到两元素之和为target public int[] twoSum(int[] nums, int target) { Map<Integer, Integer> map = new HashMap<>(); for(int i = 0; i < nums.length; i++) { if(map.containsKey(nums[i])) { return new int[]{map.get(nums[i]), i}; } map.put(target - nums[i], i); } return new int[2]; } }
标签：row inpu public shm except and contain turn new
原文地址：http://www.cnblogs.com/kevincong/p/7900352.html
