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# 題目敘述
Given an integer array nums where every element appears three times except for one, which appears exactly once. Find the single element and return it.
You must implement a solution with a linear runtime complexity and use only constant extra space.
# Example 1
Input: nums = [2,2,3,2]
Output: 3
# Example 2
Input: nums = [0,1,0,1,0,1,99]
Output: 99
# 解題思路
# Solution
class Solution { | |
public int singleNumber(int[] nums) { | |
int once = 0, twice = 0; | |
for(int num : nums){ | |
once = (num ^ once) & (~twice); | |
twice = (num ^ twice) & (~once); | |
} | |
return once; | |
} | |
} |
import java.util.HashMap; | |
import java.util.Map; | |
class Solution { | |
public int singleNumber(int[] nums) { | |
Map<Integer, Integer> mp = new HashMap<>(); | |
for(int i = 0; i < nums.length; i++){ | |
mp.put(nums[i], mp.getOrDefault(nums[i], 0) + 1); | |
if(mp.get(nums[i]) == 3){ | |
mp.remove(nums[i]); | |
} | |
} | |
int ans = 0; | |
for(int m : mp.keySet()){ | |
ans = m; | |
} | |
return ans; | |
} | |
} |
單字
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片語 & 搭配詞
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