LeetCode 1: Two Sum

最新推荐文章于 2022-12-03 16:37:21 发布

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最新推荐文章于 2022-12-03 16:37:21 发布

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文章标签： LeetCode 算法数据结构 two sum

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本文深入探讨了经典的“两数之和”问题，提供了三种不同的解决方案，包括暴力穷举、使用哈希表的空间换时间策略，以及最优化的一次遍历哈希表方法。通过这些算法的对比，读者可以理解不同技术的优劣，以及如何在实际编程中选择最合适的解决方案。

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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].

Solution1:

class Solution {
    public int[] twoSum(int[] nums, int target) {
        int[] index = new int[2];
        int firstIndex = 0;
        int secondIndex = 1;
        loop:for(firstIndex = 0;firstIndex < nums.length-1;++firstIndex){
            for(secondIndex = firstIndex+1;secondIndex < nums.length;++secondIndex){
                if(target == nums[firstIndex]+nums[secondIndex]){
                    break loop;
                }
            }
        }
        index[0] = firstIndex;
        index[1] = secondIndex;
        return index;
    }
}

非常暴力地穷举，具有 $O(n^2)$ 的时间复杂度。

Solution2: 空间换时间

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");
}

Solution3: 太机智了，只扫描一次就够了

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");
}