数据结构与算法：leetcode_01_Two Sum

最新推荐文章于 2020-04-15 13:24:08 发布

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最新推荐文章于 2020-04-15 13:24:08 发布

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分类专栏：数据结构与算法文章标签：数据结构算法 leetcode

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本文探讨了在整数数组中寻找两个数相加等于特定目标值的问题，并提供了两种解决方案。一种是简单直接的双重循环遍历方法，时间复杂度为O(n*n)；另一种利用哈希表进行优化，时间复杂度降低到O(n)。

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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].
solution_1
算法复杂度为O(n*n)

public void twoSum(int[] number, int target) {
        int length = number.length;
        for (int i = 0; i < length; i++) {
            for (int j = length - 1; j > i; j--) {
                if ((number[i] + number[j]) == target) {
                    System.out.println("those two indexes is : " + i + "," + j);
                }
            }
        }
    }

solution_2
算法复杂度为O(n)

public static void twoSumHash(int[] number,int target) {
        int length = number.length;
        HashMap<Integer,Integer> hashMap = new HashMap<>();
        for(int i=0;i<length;i++) {
            if(!hashMap.containsKey(number[i])) {
                hashMap.put(target-number[i], i);
            }else {
                System.out.println("hashmap,those two indexes is :"+hashMap.get(number[i])+","+i);
            }
        }
    }