Leetcode 1.Two Sum-蒲公英云

Leetcode 1.Two Sum

矫情吗；* 2022-09-26 02:58 226阅读 0赞

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.

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

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

解

1 穷解法。

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[j] == target - nums[i]) {
                return new int[] { i, j };
            }
        }
    }
}

没啥好说的，两层循环，时间复杂度o（n²）

2，HashMap
1.利用HashMap只能存储不能重复对象的原理，新建一个HashMap，从第一个开始循环，HashMap里面放键为target-每个数的结果值为下标.
2.然后从numbers[0]开始，然后讯问target-numbers[0]是否存在于HashMap。
3.存在就返回result，不存在就把number[0]和其下标加入HashMap

class Solution {  
    public int[] twoSum(int[] numbers, int target) {
        int[] result = new int[2];
        Map<Integer, Integer> map = new HashMap<>();
        for (int i = 0; i < numbers.length; i++) {
            if (map.containsKey(target - numbers[i])) {
                result[0] = map.get(target - numbers[i]);
                result[1] = i;
                return result;
            }
            map.put(numbers[i], i);
        }
        return result;
    }
}