1. Two Sum - Easy

标签：integer   sum   col   ber   下标   public   返回   tar   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 same element twice.

Example:

Given nums = [2, 7, 11, 15], target = 9,

Because nums[0] + nums[1] = 2 + 7 = 9,
return [0, 1].

用hashmap记录数组元素及其下标 <nums[i], i>

注意返回值要求是下标小的在前面，而从左往右遍历数组， 一般在右半部分才能找到相加为target的数，此时的index应该是较的的那个，应该放在第二位，从map中取得的index应该放在第一位。

时间：O(N)，空间：O(N)

class Solution {
    public int[] twoSum(int[] nums, int target) {
        HashMap<Integer, Integer> map = new HashMap<>();
        for(int i = 0; i < nums.length; i++) {
            if(map.containsKey(target - nums[i]))
                return new int[] {map.get(target - nums[i]), i};
            map.put(nums[i], i);
        }
        return null;
    }
}

1. Two Sum - Easy

标签：integer   sum   col   ber   下标   public   返回   tar   ash   

原文地址：https://www.cnblogs.com/fatttcat/p/10057764.html