https://leetcode.com/problems/string-to-integer-atoi/discuss/
实现 atoi 函数
思路:
循环按位解析,需要手动判断溢出等情况,写的有点乱,之后再精简下.
class Solution {
public int myAtoi(String str) {
str = str.trim();
if(str.length() == 0){
return 0;
}
int number=0;
int fu=1;
int startIndex=0;
if(str.charAt(0)=='-'){
fu=-1;
startIndex=1;
}else if(str.charAt(0)=='+'){
startIndex=1;
}
for(int i=startIndex;i<str.length();i++){
char c=str.charAt(i);
if(isNumber(c)){
int n=c-'0';
if(fu==1 && ( number*10>=(Integer.MAX_VALUE-n) || number>Integer.MAX_VALUE/10 )){
number=Integer.MAX_VALUE;
return number;
}else if(fu==-1 && (-10*number<=(Integer.MIN_VALUE+n)|| number>Integer.MIN_VALUE/-10 )){
number=Integer.MIN_VALUE;
return number;
}
number=number*10+n;
}else{
break;
}
}
return fu*number;
}
private boolean isNumber(char c){
return c>='0'&&c<='9';
}
}
