class Solution: def sumOfNumberAndReverse(self, num: int) -> bool: # 枚举 [0, num] 内的每个数 for i in range(num + 1): # 如果当前 i 满足题意,则直接返回 true if i + Solution.reverse(i) == num: return True # 此时表明无满足题意的数,直接返回 false return False @staticmethod d...
publicclassSolution {publicvoidsolveSudoku(char[][] board) {if(board ==null|| board.length == 0)return; solve(board); }publicbooleansolve(char[][] board){for(inti = 0; i < board.length; i++){for(intj = 0; j < board[0].length; j++){if(board[i][j] == '.'){for(charc...
public int reverse(int x) { int ret=0; while(x!=0) { int t=x%10; ret=ret*10+t; x=x/10; } return ret; } } publicclassSolution {publicbooleanisPalindrome(intx) {if(x<0)returnfalse;if(x==0)returntrue;intret=0;intxold=x;while(x!=0) {inttemp= x%10;//zui hou yi wei...
leetcode:Reverse Integer 及Palindrome Number Reverse IntegerReverse digits of an integer.Example1: x = 123, return 321Example2: x = -123, return -321click to show spoilers.Have you thought about this?Here are some good questions to ask before coding. Bonus points for you if you have ...
View Code SOLUTION 3 使用一个专用的反转函数来进行反转,从头到尾遍历,遍历到K的时候,使用Pre-Next指针的方式进行反转。这个方法比递归更棒。 要特别注意的是: reverseSection 函数中,while 循环的终止条件不是cur != null,而是cur != next。这一点要特别注意,否则很容易造成死循环!
Solution: https://leetcode.com/problems/reverse-integer/discuss/229800/Python3-Faster-than-100 classSolution:defreverse(self, x):""":type x: int :rtype: int"""label= 1ifx<0: label= -1s=str(abs(x)) S= label*int(s[-1:None:-1])ifS<-2**31orS>2**31-1:return0else:returnS ...
leetcode题解 7.Reverse Integer 题目: Given a 32-bit signed integer, reverse digits of an integer. Example 1: Input:123Output:321 Example 2: Input:-123Output:-321 Example 3: Input:120Output:21 Note: Assume we are dealing with an environment which could only hold integers within the 32-...
leetcode/discuss思路: 思路1:新建一个uint32_t的数r:a:r <<=1;r每次左移,则原二进制数的最低位就变成新数的最高位了 b:r |=n &1;如果原数值为1则为1,为0则为0 (可否看做字符串,不可以!!因为输入的为整型数) c:n >>= 1;原数值每次右移1位 ...
Reverse digits of an integer. Example1: x = 123, return 321 Example2: x = -123, return -321 2.解法分析 解法需注意: 为零的状况 为负数的状况 溢出怎么处理 结尾一串0怎么处理 代码如下: classSolution { public: intreverse(intx) {
则最后栈里面剩下的元素即为所求。 [Code] 1: intevalRPN(vector<string> &tokens) {2: stack<int> operand;3:for(int i =0;i< tokens.size();i++)4: {5: if ((tokens[i][0] =='-'&& tokens[i].size()>1) //negative number6: || (tokens[i][0] >='0'&& tokens[i][0] <=...