LeetCode-String to Integer (atoi)
2015-06-03 05:48
399 查看
corner case很多需要注意
1.这个题规定前后的white space都忽略掉 可以有正负号 但是后面就不可以有特殊符号了 比如 +182@#¥ 就返回+182
2. 注意得到数值 是用 c - '0'
3.超int范围的 min max 尤其是负数超范围 因为最开始先把符号存起来 当正数对待了 所以要在过程中判断是否超了max 但是是否是正的min
public class Solution {
public int myAtoi(String str) {
if (str == null || str.length() == 0)
return 0;
str = str.trim();
boolean neg = false;
int index = 0;
if ( str.charAt(0) == '-' ){
neg = true;
index ++;
}
if ( str.charAt(0) == '+' ){
index ++;
}
long ans = 0;
while ( index < str.length() ){
if ( str.charAt(index) -'0' >= 0 && str.charAt(index) -'0' <= 9){
ans = ans * 10 + str.charAt(index) -'0';
System.out.println(ans);
if ( ans >= Integer.MAX_VALUE && !neg){
ans = Integer.MAX_VALUE;
System.out.println("what");
break;
}
if ((0 - ans) <= Integer.MIN_VALUE && neg){
return Integer.MIN_VALUE;
}
index ++;
}
else
break;
}
if ( neg )
ans = 0 - ans;
return (int) ans;
}
}
1.这个题规定前后的white space都忽略掉 可以有正负号 但是后面就不可以有特殊符号了 比如 +182@#¥ 就返回+182
2. 注意得到数值 是用 c - '0'
3.超int范围的 min max 尤其是负数超范围 因为最开始先把符号存起来 当正数对待了 所以要在过程中判断是否超了max 但是是否是正的min
public class Solution {
public int myAtoi(String str) {
if (str == null || str.length() == 0)
return 0;
str = str.trim();
boolean neg = false;
int index = 0;
if ( str.charAt(0) == '-' ){
neg = true;
index ++;
}
if ( str.charAt(0) == '+' ){
index ++;
}
long ans = 0;
while ( index < str.length() ){
if ( str.charAt(index) -'0' >= 0 && str.charAt(index) -'0' <= 9){
ans = ans * 10 + str.charAt(index) -'0';
System.out.println(ans);
if ( ans >= Integer.MAX_VALUE && !neg){
ans = Integer.MAX_VALUE;
System.out.println("what");
break;
}
if ((0 - ans) <= Integer.MIN_VALUE && neg){
return Integer.MIN_VALUE;
}
index ++;
}
else
break;
}
if ( neg )
ans = 0 - ans;
return (int) ans;
}
}
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