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  • [leetcode]String to Integer (atoi)

    String to Integer (atoi)

     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.

    算法思路:

    坑比较多,没难度。开整之前建议先把str预处理一下

    【注意】边界case:

    全空格、全非法字符、空串<-这三种预处理之后都变成空串了。

    溢出。因此先把result声明成long型。

    符号处理:

    两个boolean变量表示正负号,但是一旦有了某个符号,第二个符号就变成非法字符了。字符串中未出现符号,按照正号处理。

    代码如下:

     1 public class Solution {
     2     public int atoi(String str) {
     3         if(str == null || str.trim().length() == 0) return 0;
     4         str = str.trim();
     5         StringBuilder sb = new StringBuilder();
     6         int length = str.length();
     7         ////preprocess,only maintain num and operater
     8         for(int i = 0; i < length; i++){
     9             if(str.charAt(i) == '+' || str.charAt(i) == '-'  || (str.charAt(i) <= '9' && str.charAt(i) >= '0')){
    10                 sb.append(str.charAt(i));
    11             }else break;
    12         }
    13         length = sb.length();
    14         boolean positive = false,negative = false;
    15         long res = 0;
    16         for(int i = 0; i < length; i++){
    17             //the second time appearrance of operater is invalide
    18             if((sb.charAt(i) == '+' || sb.charAt(i) == '-') && (positive || negative)) 
    19                 break;
    20             if(sb.charAt(i) == '+'){
    21                 positive = true;
    22             }else if(sb.charAt(i) == '-'){
    23                 negative = true;
    24             }else{
    25                 res = res * 10 + (sb.charAt(i) - '0');
    26             }
    27         }
    28         if(!positive && !negative) positive = true;
    29         // process overflow situation
    30         if(positive) return res > Integer.MAX_VALUE ? Integer.MAX_VALUE : (int) res;
    31         return res > Integer.MAX_VALUE ? Integer.MIN_VALUE : (int)res * -1;        
    32     }
    33 }
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  • 原文地址:https://www.cnblogs.com/huntfor/p/3911519.html
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