首页 > 其他分享 >leetcode-202-easy

leetcode-202-easy

时间:2022-10-19 07:47:57浏览次数:56  
标签:10 202 return int sum set easy leetcode

Happy Number
思路一: happy number 的结果完全分类,就两种情况

  1. 最后的值为 1
  2. 进入循环(用 map 记录)
public boolean isHappy(int n) {
    Set<Integer> set = new HashSet<>();
    while (true) {

        int sum = 0;
        while (n >= 1) {
            sum += (n % 10) * (n % 10);
            n /= 10;
        }
        if (sum == 1) return true;
        n = sum;

        if (set.contains(sum)) {
            return false;
        } else {
            set.add(sum);
        }
    }
}

标签:10,202,return,int,sum,set,easy,leetcode
From: https://www.cnblogs.com/iyiluo/p/16804870.html

相关文章

  • leetcode-141-easy
    LinkedListCycle思路一:用set记录每个节点的hashCode,如果遇到重复,说明是循环publicbooleanhasCycle(ListNodehead){Set<Integer>set=newHashSet<>();......
  • leetcode-344-easy
    ReverseString思路一:首尾互换publicvoidreverseString(char[]s){intmid=s.length/2;intbegin=0;intend=s.length-1;while(beg......
  • leetcode-263-easy
    UglyNumber思路一:从数字中依次去除2,3,5,查看剩余的值是否为1publicbooleanisUgly(intn){if(n==0)returnfalse;while(n%2==0){n/......
  • leetcode-217-easy
    ContainsDuplicate思路一:Set检测publicstaticbooleancontainsDuplicate(int[]nums){Set<Integer>set=newHashSet<>();for(intnum:nums){......
  • leetcode-709-easy
    ToLowerCase思路一:遍历,遇到A-Z范围内的char转换到对应a-z范围publicStringtoLowerCase(Strings){if(s==null||s.isEmpty())returns;int......
  • leetcode-415-easy
    AddString思路一:模拟加法运算,字符串前面填零publicStringaddStrings(Stringnum1,Stringnum2){intmax=Math.max(num1.length(),num2.length());nu......
  • leetcode-389-easy
    FindtheDifference思路一:xor两个字符串publiccharfindTheDifference(Strings,Stringt){charresult=0;for(inti=0;i<s.length();i++){......
  • leetcode-226-easy
    InvertBinaryTree思路一:递归,交换左右。这题比较出名,个人感觉面试的题目和实际工作中遇到的问题还是不太一样的,所以一点准备都不做就跑去面试,答不上来很正常。一般能力......
  • leetcode-219-easy
    ContainsDuplicateII思路一:for循环遍历,结果超时publicbooleancontainsNearbyDuplicate(int[]nums,intk){intleft=-1;for(inti=0;i<nums......
  • leetcode-448-easy
    思路一:用数组记录publicList<Integer>findDisappearedNumbers(int[]nums){int[]m=newint[nums.length];for(intnum:nums){if(num-1......