LeetCode——String to Integer (atoi)
来源:互联网 发布:越南古代服饰淘宝 编辑:程序博客网 时间:2024/06/05 23:50
Implement atoi to convert a string to an integer.
Hint: Carefully consider all possible input cases. If you want a challenge, please do not see below and ask yourself what are the possible input cases.
Notes: It is intended for this problem to be specified vaguely (ie, no given input specs). You are responsible to gather all the input requirements up front.
spoilers alert... click to show requirements for atoi.
The function first discards as many whitespace characters as necessary until the first non-whitespace character is found. Then, starting from this character, takes an optional initial plus or minus sign followed by as many numerical digits as possible, and interprets them as a numerical value.
The string can contain additional characters after those that form the integral number, which are ignored and have no effect on the behavior of this function.
If the first sequence of non-whitespace characters in str is not a valid integral number, or if no such sequence exists because either str is empty or it contains only whitespace characters, no conversion is performed.
If no valid conversion could be performed, a zero value is returned. If the correct value is out of the range of representable values, INT_MAX (2147483647) or INT_MIN (-2147483648) is returned.
实现 atoi 将字符串转为整数。
需要考虑正负号,空格,非数字字符出现的话,其后部分随之忽略。
提交了如下的代码,结果运行超时:
public static int atoi(String str) {str = str.trim();if(str.isEmpty())return 0;char[] ch = str.toCharArray();for(int i=1;i<ch.length;i++){if(!Character.isDigit(ch[i])){str = str.substring(0, i);break;}}ch = str.toCharArray();int ret = 0;for (int i = 0; i < ch.length; i++) {if (ch[0] == '-') {ret = ret * 10 - (ch[i + 1] - 48);if (i == ch.length - 2)break;} else if (ch[0] == '+') {ret = ret * 10 + (ch[i + 1] - 48);if (i == ch.length - 2)break;} else {ret = ret * 10 + (ch[i] - 48);}}return ret;}
下面是抄了网友http://my.oschina.net/jdflyfly/blog/283516的代码来学习一下:
public static int atoi(String str) {int max = Integer.MAX_VALUE;int min = -Integer.MIN_VALUE;long result = 0;str = str.trim();int len = str.length();if (len < 1)return 0;int start = 0;boolean neg = false;if (str.charAt(start) == '-' || str.charAt(start) == '+') {if (str.charAt(start) == '-')neg = true;start++;}for (int i = start; i < len; i++) {char ch = str.charAt(i);if (ch < '0' || ch > '9')break;result = 10 * result + (ch - '0');if (!neg && result > max)return max;if (neg && -result < min)return min;}if (neg)result = -result;return (int) result;}
- LeetCode——String to Integer (atoi)
- LeetCode——String to Integer (atoi)
- String to Integer (atoi) —— Leetcode
- leetcode——String to Integer (atoi)
- LeetCode—8. String to Integer (atoi)
- leetcode——8——String to Integer (atoi)
- leetcode 008 —— String to Integer (atoi)
- LeetCode题解——8 String to Integer (atoi)
- #8 LeetCode——String to Integer (atoi)
- leetcode第8题——*String to Integer (atoi)
- LeetCode之路——String to Integer (atoi)
- LeetCode算法题——String to Integer (atoi)
- 刷LeetCode(8)——String to Integer (atoi)
- [LeetCode]—String to Integer (atoi) 字符串转换为数字
- Leetcode Problem.8—String to Integer (atoi)
- LeetCode 之 String to Integer (atoi) — C 实现
- LeetCode:String to Integer (atoi)
- LeetCode: String to Integer (atoi)
- 学习总结十八:Hbase数据库介绍
- PowerManager之UserActivity
- Java递归搜索指定文件夹下的匹配文件
- 温度PID控制
- 新生报到
- LeetCode——String to Integer (atoi)
- double long float类型读入读出 double取模 fmod
- 【LeetCode with Python】 Binary Tree Level Order Traversal II
- Cocos2d-iPhone V3 (2) 场景转换
- PHP环境搭建:Windows7系统配置PHP+Apache+MySQL环境教程
- (集合)JAVA文件操作--复制、粘贴、删除、剪切
- 【Nutch2.2.1基础教程之2.1】集成Nutch/Hbase/Solr构建搜索引擎之一:安装及运行【单机环境】
- 人工智能传奇——关于AI起源与发展的故事
- OpenVPN多实例优化的思考过程