Leetcode 1. Two Sum, Java 解法

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].

Approach #1 Brute Force 我的解法

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

• Time complexity : O(n^2). For each element, we try to find its complement by looping through the rest of array which takes O(n) time. Therefore, the time complexity is O(n^2).

• Space complexity : O(1). 

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");
}

• Time complexity : O(n)O(n). We traverse the list containing nn elements exactly twice. Since the hash table reduces the look up time to O(1)O(1), the time complexity is O(n)O(n).

• Space complexity : O(n)O(n). The extra space required depends on the number of items stored in the hash table, which stores exactly nn elements. 

通过牺牲一部分内存空间来换取更快的运行速度，使用Hash Map来完成任务, 因为哈希表的查找元素时间接近O(1). 用到了map.get(), map.put() 这些之前用过的方法。

Approach #3 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");

• Time complexity : O(n). We traverse the list containing nn elements only once. Each look up in the table costs only O(1) time.

• Space complexity : O(n) The extra space required depends on the number of items stored in the hash table, which stores at most nn elements.

在进行循环并插入元素到表的过程中，同时检查complement 是否已经在表中，如果有解，直接返回。