260. Single Number III

Posted 2021-03-25 wentiliangkaihua

Given an array of numbers nums, in which exactly two elements appear only once and all the other elements appear exactly twice. Find the two elements that appear only once.

Example:

Input:  [1,2,1,3,2,5]
Output: [3,5]

Note:

1. The order of the result is not important. So in the above example, [5, 3] is also correct.
2. Your algorithm should run in linear runtime complexity. Could you implement it using only constant space complexity?

class Solution {
    public int[] singleNumber(int[] nums) {
        List<Integer> list = singleNumber1(nums);
        int[] res = new int[list.size()];
        for(int i = 0; i < list.size(); i++){
            res[i] = list.get(i);
        }
        return res;
    }
    
        public List<Integer> singleNumber1(int[] nums) {
        List<Integer> list = new ArrayList();
        Map<Integer, Integer> map = new HashMap();
        for(int i : nums){
            int value = map.getOrDefault(i, 0);
            map.put(i, value + 1);
        }
        for(Map.Entry<Integer, Integer> entry : map.entrySet()){
            if(entry.getValue() == 1) list.add(entry.getKey());
        }
        return list;
    }
}

调用1中map方法，屡试不爽

class Solution {
    public int[] singleNumber(int[] nums) {
    if(nums == null || nums.length == 0) return new int[0];
    
    Map<Integer, Integer> map = new HashMap<>();
    
    for(int i = 0; i < nums.length; i++)
    {
        if(!map.containsKey(nums[i]))
        {
            map.put(nums[i], 1);
        }else{
            map.remove(nums[i]);
        }
    }
    
    int[] output = new int[2];
    
    int n = 0;
    
    for(int i : map.keySet())
    {
        output[n++] = i;
    }
    
    return output;
    
}
    }

或者稍微优化一下。