1. Two Sum

最新推荐文章于 2024-09-07 16:05:05 发布

原创最新推荐文章于 2024-09-07 16:05:05 发布 · 260 阅读

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本文详细介绍了LeetCode上经典问题“两数之和”的三种解法：暴力法、两次哈希表法及一次哈希表法。通过对比不同方法的时间复杂度与空间复杂度，帮助读者理解并掌握高效解决问题的技巧。

题目描述

在这里插入图片描述

题目链接

https://leetcode.com/problems/two-sum/

方法思路

Approach1：

public int[] twoSum(int[] nums, int target) {
        for(int i = 0; i < nums.length; i++){
            for(int j = nums.length - 1; j > i; j--){
                if(nums[i] + nums[j] == target){
                    return new int[]{i, j};
                }
            }
        }
        return new int[2];
    }

Approach2：Two-pass Hash Table

public int[] twoSum(int[] nums, int target) {
    Map<Integer, Integer> map = new HashMap<>();
    for (int i = 0; i < nums.length; i++) {
        map.put(nums[i], i);
    }
    for (int i = 0; i < nums.length; i++) {
        int complement = target - nums[i];
        if (map.containsKey(complement) && map.get(complement) != i) {
            return new int[] { i, map.get(complement) };
        }
    }
    throw new IllegalArgumentException("No two sum solution");
}

Approach3：One-pass Hash Table

public int[] twoSum(int[] nums, int target) {
    Map<Integer, Integer> map = new HashMap<>();
    for (int i = 0; i < nums.length; i++) {
        int complement = target - nums[i];
        if (map.containsKey(complement)) {
            return new int[] { map.get(complement), i };
        }
        map.put(nums[i], i);
    }
    throw new IllegalArgumentException("No two sum solution");
}