leetcode 1. Two Sum-优快云博客

本文链接：https://blog.youkuaiyun.com/OliverZang/article/details/86626037

本文探讨了LeetCode题目“两数之和”的多种解决方案，包括暴力破解法和使用哈希表的两遍及一遍解法。通过对比不同方法的运行效率，展示了哈希表在解决此类问题时的优势。

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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].

My solution: brute force

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

test result:

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Runtime: 33 ms, faster than 28.56% of Java online submissions for Two Sum.

The below solution is provided by leetcode solutions:

Approach 2: 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");
}

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");
}

Summary:

The first attempt to the leetcode Q.No.1 succeed with pretty low efficiency.
I just felt a bit depressed ? (kidding)
The other two solutions has amazed me by its running speed. Somehow using hashmap
is kind of cheating but is a smart solution.