Implement the myAtoi(string s) function, which converts a string to a 32-bit signed integer (similar to C/C++'s atoi function).
The algorithm for myAtoi(string s) is as follows:
'-' or '+'. Read this character in if it is either. This determines if the final result is negative or positive respectively. Assume the result is positive if neither is present."123" -> 123, "0032" -> 32). If no digits were read, then the integer is 0. Change the sign as necessary (from step 2).[-231, 231 - 1], then clamp the integer so that it remains in the range. Specifically, integers less than -231 should be clamped to -231, and integers greater than 231 - 1 should be clamped to 231 - 1.Note:
' ' is considered a whitespace character.Example 1:
Input: s = "42"
Output: 42
Explanation: The underlined characters are what is read in, the caret is the current reader position.
Step 1: "42" (no characters read because there is no leading whitespace)
^
Step 2: "42" (no characters read because there is neither a '-' nor '+')
^
Step 3: "42" ("42" is read in)
^
The parsed integer is 42.
Since 42 is in the range [-231, 231 - 1], the final result is 42.
Example 2:
Input: s = " -42"
Output: -42
Explanation:
Step 1: " -42" (leading whitespace is read and ignored)
^
Step 2: " -42" ('-' is read, so the result should be negative)
^
Step 3: " -42" ("42" is read in)
^
The parsed integer is -42.
Since -42 is in the range [-231, 231 - 1], the final result is -42.
Example 3:
Input: s = "4193 with words"
Output: 4193
Explanation:
Step 1: "4193 with words" (no characters read because there is no leading whitespace)
^
Step 2: "4193 with words" (no characters read because there is neither a '-' nor '+')
^
Step 3: "4193 with words" ("4193" is read in; reading stops because the next character is a non-digit)
^
The parsed integer is 4193.
Since 4193 is in the range [-231, 231 - 1], the final result is 4193.
- class Solution {
- public int myAtoi(String str) {
- if (str == null || str.length() == 0){
- return 0;
- }
-
- int index = 0;
-
- while(index < str.length() && str.charAt(index) == ' '){
- index++; // 这一步去掉空格
- }
-
- if(index == str.length()){
- return 0; //去掉完以后发现string没了,直接return
- }
-
- int sign = 1;
- if(str.charAt(index) == '+' || str.charAt(index) == '-'){
- sign = str.charAt(index) == '+' ? 1 : -1;
- index++; //这一步收集正负号,到时候乘以sign即可
- }
-
- int res = 0;
- while(index < str.length()){
- int digit = str.charAt(index) - '0';
- if(digit < 0 || digit > 9) break; //如果不是数字而是字符的话就返回已有的东西
- int newRes = res * 10 + digit;
- if(newRes / 10 != res){ //这一步是防止溢出
- return sign == 1 ? Integer.MAX_VALUE : Integer.MIN_VALUE;
- }
- res = newRes;
- index ++;
- }
-
- return res * sign; //最后别忘了乘以符号
- }
- }