zoukankan      html  css  js  c++  java
  • Leetcode -- Day 51 & Day 55

    String to Integer (atoi) / valid number

    Question 1

    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.

     1 public int myAtoi(String str) {
     2         if (str == null || str.length() < 1){
     3             return 0;
     4         }
     5         str = str.trim();
     6         boolean isNeg = false;
     7         int i = 0;
     8         
     9         if (str.charAt(0) == '-'){
    10             isNeg = true;
    11             str = str.substring(1);
    12         }
    13         else if (str.charAt(0) == '+'){
    14             str = str.substring(1);
    15         }
    16         
    17         if (!Character.isDigit(str.charAt(0))){
    18             return 0;
    19         }
    20         
    21         double result = 0;
    22         for (char c : str.toCharArray()){
    23             if (Character.isDigit(c))
    24                 result = result * 10 + (c-'0');
    25             else 
    26                 break;
    27         }
    28         if (isNeg){
    29             result = - result;
    30             return result < Integer.MIN_VALUE ? Integer.MIN_VALUE : (int) result;
    31         }
    32         else{
    33             return result > Integer.MAX_VALUE ? Integer.MAX_VALUE : (int) result;
    34         }
    35     }

    Question 2

    Valid Number

    Validate if a given string is numeric.

    Some examples:
    "0" => true
    " 0.1 " => true
    "abc" => false
    "1 a" => false
    "2e10" => true

    Note: It is intended for the problem statement to be ambiguous. You should gather all requirements up front before implementing one.

    For this problem, wee need to set eIndex and dotIndex to record the index of char e and . , which is very important for the special valid number cases.

     1     public boolean isNumber(String s) {
     2         s = s.trim();
     3         int eIndex = -1;
     4         int dotIndex = -1;
     5         
     6         if (s.length()==0)
     7             return false;
     8         
     9         int i = 0;
    10         
    11         if (s.charAt(i) == '+' || s.charAt(i) == '-')
    12             s = s.substring(1);
    13             
    14         for (i = 0; i < s.length(); i ++){
    15             if (eIndex == -1 && s.charAt(i) == 'e'){
    16                 eIndex = i;
    17                 if (i + 1 < s.length() && (s.charAt(i+1) == '-' || s.charAt(i+1) == '+'))
    18                     i ++;
    19             }
    20             else if (dotIndex == -1 && s.charAt(i) == '.')
    21                 dotIndex = i;
    22             else{
    23                 if (Character.isDigit(s.charAt(i)))
    24                     continue;
    25                 else
    26                     return false;
    27             }
    28         }
    29         
    30         String startString, midString, endString;
    31         if (eIndex == -1 && dotIndex == -1){
    32             startString = s;
    33             if (startString.length() < 1)
    34                 return false;
    35         }
    36         else if (eIndex == -1 && dotIndex != -1){
    37             startString = s.substring(0, dotIndex);
    38             endString = s.substring(dotIndex+1);
    39             if (startString.length()<1 && endString.length() < 1)
    40                 return false;
    41         }
    42         else if (dotIndex == -1 && eIndex != -1){
    43             startString = s.substring(0, eIndex);
    44             if (startString.length()<1)
    45                 return false;
    46             if (eIndex+1 < s.length() && (s.charAt(eIndex+1) == '+' ||s.charAt(eIndex+1) == '-'))
    47                 endString = s.substring(eIndex + 2);
    48             else
    49                 endString = s.substring(eIndex + 1);
    50             if (endString.length()<1)
    51                 return false;
    52         }
    53         else{
    54             if (dotIndex > eIndex)  return false;
    55             else{
    56                 startString = s.substring(0, dotIndex);
    57                 midString = s.substring(dotIndex+1, eIndex);
    58                 if (startString.length()<1 && midString.length()<1)
    59                     return false;
    60                 if (eIndex+1 < s.length() && (s.charAt(eIndex+1) == '+' ||s.charAt(eIndex+1) == '-'))
    61                     endString = s.substring(eIndex + 2);
    62                 else
    63                     endString = s.substring(eIndex + 1);
    64                 if (endString.length()<1 )
    65                     return false;
    66             }
    67         }
    68         return true;
    69     }
  • 相关阅读:
    原型链
    原型规则总结
    使用typeof能得到的哪些类型
    eslint 规则中文注释
    两张图片互相切换
    输入框获得焦点与失去焦点、阴影效果
    vue 写table的几种方式
    vue 注意事项
    angular 接口定义封装
    @NgModule
  • 原文地址:https://www.cnblogs.com/timoBlog/p/4748144.html
Copyright © 2011-2022 走看看