日期：20180912
題目描述：
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 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.
Note:
- Only the space character
' 'is considered as whitespace character. - Assume we are dealing with an environment which could only store integers within the 32-bit signed integer range: [?231, 231 ? 1]. If the numerical value is out of the range of representable values, INT_MAX (231 ? 1) or INT_MIN (?231) is returned.
Example 1:
Input: "42"
Output: 42
Example 2:
Input: " -42"
Output: -42
Explanation: The first non-whitespace character is '-', which is the minus sign.
Then take as many numerical digits as possible, which gets 42.
Example 3:
Input: "4193 with words"
Output: 4193
Explanation: Conversion stops at digit '3' as the next character is not a numerical digit.
Example 4:
Input: "words and 987"
Output: 0
Explanation: The first non-whitespace character is 'w', which is not a numerical
digit or a +/- sign. Therefore no valid conversion could be performed.
Example 5:
Input: "-91283472332"
Output: -2147483648
Explanation: The number "-91283472332" is out of the range of a 32-bit signed integer.
Thefore INT_MIN (?231) is returned.
詳解：
這次居然一次寫完直接通過，都沒有debug，還擊敗了96.78%的人，于是我得出一個(gè)結(jié)論：
經(jīng)過了這么久的練習(xí)，我終于
。
。
。
。
。
。
遇到了一道簡單的題。
class Solution {
public:
int myAtoi(string str) {
long long int x = 0,tx = 0;
bool isneg = false;
int i = 0;
while(str[i]==' '){
i++;
}
if(str[i]=='-'||str[i]=='+'||(str[i]>='0'&&str[i]<='9')){
if(str[i]=='-'){
isneg = true;
i++;
}else if(str[i]=='+'){
i++;
}
while(str[i]>='0'&&str[i]<='9'){
x *= 10;
x += str[i] - '0';
if(isneg){
tx = -x;
}else{
tx = x;
}
if(tx>2147483647){
return 2147483647;
}
if(tx<-2147483648){
return -2147483648;
}
i++;
}
int res = tx;
return res;
}else{
return 0;
}
}
};
做法和簡單粗暴，依然用long long int來保存真實(shí)值，在寫代碼的時(shí)候確實(shí)可以簡單粗暴的不停的if，即把所有的情況一一寫出來，這樣雖然代碼很長，但是運(yùn)行速度不會(huì)慢，甚至更快，而且可讀性一般也不會(huì)差。
高度凝練的代碼需要千錘百煉才能寫出，筆試時(shí)往往受時(shí)間的限制不適合這么做，并且高度凝練的代碼往往可讀性不會(huì)太好。
