实现代码: classSolution{ // Simulation // N is the length of s // Time Complexity: O(N) // Space Complexity: O(1) publicintmyAtoi(String s){ if(s.isEmpty())return0; // 去除前导空格,满足步骤 1 intstart =0; char[] str = s.toCharArray(); while(start < str.length && str[s...
String to Integer (atoi) - LeetCode 注意点 溢出时返回上界或下界 解法 解法一:从前往后,首先跳过空白字符,然后对非空白字符进行判断,如果是数字字符就转化为数字,否则就返回结果。过程中随时判断是否溢出。时间复杂度为O(n) class Solution { public: int myAtoi(string str) { int i = 0; bool positive...
Can you solve this real interview question? String to Integer (atoi) - Implement the myAtoi(string s) function, which converts a string to a 32-bit signed integer. The algorithm for myAtoi(string s) is as follows: 1. Whitespace: Ignore any leading whi
classSolution{public:intmyAtoi(strings){if(s.empty()){return0;}inti=0,len=s.length();intres=0,firstChar=1;// // firstChar 表示正负号,默认为 '+',即 1intupperBound=INT_MAX/10;while(s[i]==' '){// 省去所有空格if(++i>=len){return0;}}if(s[i]=='-'){// 判断正负号firstCh...
* @链接:https://leetcode-cn.com/problems/string-to-integer-atoi/solution/javascripttou-ji-qu-qiao-wu-xu-si-kao-yi-kan-jiu-h/ * @param {string} str * @return {number} */varmyAtoi=function(str){constnumber=parseInt(str,10);if(isNaN(number)){return0;}elseif(number<Math.pow(-2,...
leetcode 8. String to Integer (atoi) 手写字符串转整数函数。细节好多,首先要处理掉前导空格,然后处理正负号,接下来往后处理时,处理到非数字字符时,立即结束返回前面得到的整数,最后如果溢出的话,返回离它最近的一个整数 class Solution(object): def myAtoi(self, str):...
class Solution { public: int myAtoi(string str) { if(str.empty()) {return 0;} int i = 0; int sign = 1; int sum = 0; int n = str.length(); while(i < n && ' ' == str[i]) { ++i; } if('+' == str[i]) { ++i; } else if('-' == str[i]) { ++i; sign ...
Implement themyAtoi(string s)function, which converts a string to a 32-bit signed integer. The algorithm formyAtoi(string s)is as follows: Whitespace: Ignore any leading whitespace (" "). Signedness: Determine the sign by checking if the next character is'-'or'+', assuming positivity if...
今天的题目是(8. String to Integer (atoi))[https://leetcode.com/problems...] 题目描述: Implement atoi which converts a string to an integer. The function first discards as many whitespace characters as necessary until the first non-whitespace character is found. Then, starting from this char...
return Integer.MIN_VALUE; return (int) result; } public static void main(String[] args) { Solution solution=new Solution(); System.out.println(solution.myAtoi("2147483648")); } } 1. 2. 3. 4. 5. 6. 7. 8. 9. 10. 11.