letcode刷题--two sum

题目描述：

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].

first:暴力法：  O(n2)

class Solution {
    public int[] twoSum(int[] nums, int target) {
        int i = 0;
        int j = 0;
        
        for(i=0; i<nums.length; i++){
            for(j=i+1; j<nums.length;j++){
                if((nums[i]+nums[j])==target){
                    if(i!=j){  
                        return new int[] { i, j };
                    }else{
                        ;
                    }
                }
            }
        }
        throw new IllegalArgumentException("No two sum solution");
    }
}

第二种方法：双通哈希表O(n)

在我认为这个算法的时间复杂度依旧为O(n2)

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");
}

3：单通哈希表

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);//添加到map中

//相当于（2，1），
//（3，（2，1）），
//（4，（3，2，1），
    }
    throw new IllegalArgumentException("No two sum solution");

 第三种最聪明