C++代码如下: classSolution{public:intlongestPalindrome(string s){ unordered_map<char,int> count;for(charc : s) ++count[c];intres =0;boolhasOne =false;for(autod : count) {if(d.second %2==0) res += d.second;else{ res += d.second -1; hasOne =true; } }if(hasOne) ++res;ret...
1classSolution(object):2defisPalindrome(self, x):3"""4:type x: int5:rtype: bool6"""7x2 = str(x)8ifx2 == x2[::-1]:9returnTrue10else:11returnFalse 一个比较精简的代码 运行时间打败了97%的代码 但是很占内存
来源:力扣(LeetCode) 著作权归作者所有。商业转载请联系作者获得授权,非商业转载请注明出处。public class Solution { public bool IsPalindrome(int x) { // 特殊情况: // 如上所述,当 x < 0 时,x 不是回文数。 // 同样地,如果数字的最后一位是 0,为了使该数字为回文, ...
class Solution: def romanToInt(self,rnum): rnum = input("input:") dic = {"I":1,"V":5,"X":10,"L":50,"C":100,"D":500,"M":1000} rnum = list(rnum) s =0 if len(rnum) <1 or len(rnum)> 15: return 0 for i in range(len(rnum)-1): if dic[rnum[i]] < dic...
最长回文串构造完成。 观察上面的构造过程,可以发现最后会剩余若干个出现奇数次的字符。那么答案就变成了统计有多少种字符出现了奇数次,我们可以用哈希来解决这个问题,如下图: 这样我们只需遍历一次 hash 数组,统计奇数的个数,然后做一次减法即可。 class Solution { public: int longestPalindrome(string s) { int...
对于循环判断的条件是不是用low<=high更好，避免每次都调用string.size()。
# self.next=NoneclassSolution(object):defisPalindrome(self,head):""":type head:ListNode:rtype:bool""" node_list=[]whilehead:node_list.append(head)head=head.nextforiinrange(int(len(node_list)/2)):print(node_list[i].val,node_list[-i-1].val)ifnode_list[i].val!=node_list[-i-1...
classSolution(object):deflongestPalindrome(self, s):""" :type s: str :rtype: int """ans = odd =0cnt = collections.Counter(s)forcincnt: ans += cnt[c]ifcnt[c] %2==1: ans -=1odd +=1returnans + (odd >0) 1 2 3
通过改变常数0x5f375a86 和1.5可以控制精度。代码虽短,学问很大,推荐看<<算法心得,高效算法的奥秘>>,很多精髓。 进入今天第二个主题,LeetCode 214 : Given a string S, you are allowed to convert(转换) it to a palindrome(回文) by adding characters in front of it. Find and return the shortestpalindr...
[leetcode] 1616. Split Two Strings to Make Palindrome Description You are given two strings a and b of the same length. Choose an index and split both strings at the same index, splitting a into two strings: aprefix and asuffix where a = aprefix + asuffix, and splitting b into two...