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  • 1. Two Sum

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