代码拉取完成,页面将自动刷新
同步操作将从 doocs/leetcode 强制同步,此操作会覆盖自 Fork 仓库以来所做的任何修改,且无法恢复!!!
确定后同步将在后台操作,完成时将刷新页面,请耐心等待。
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.
Example 4:
Input: s = "words and 987" Output: 0 Explanation: Step 1: "words and 987" (no characters read because there is no leading whitespace) ^ Step 2: "words and 987" (no characters read because there is neither a '-' nor '+') ^ Step 3: "words and 987" (reading stops immediately because there is a non-digit 'w') ^ The parsed integer is 0 because no digits were read. Since 0 is in the range [-231, 231 - 1], the final result is 0.
Example 5:
Input: s = "-91283472332"
Output: -2147483648
Explanation:
Step 1: "-91283472332" (no characters read because there is no leading whitespace)
^
Step 2: "-91283472332" ('-' is read, so the result should be negative)
^
Step 3: "-91283472332" ("91283472332" is read in)
^
The parsed integer is -91283472332.
Since -91283472332 is less than the lower bound of the range [-231, 231 - 1], the final result is clamped to -231 = -2147483648.
Constraints:
0 <= s.length <= 200
s
consists of English letters (lower-case and upper-case), digits (0-9
), ' '
, '+'
, '-'
, and '.'
.class Solution:
def myAtoi(self, s: str) -> int:
if not s:
return 0
n = len(s)
if n == 0:
return 0
i = 0
while s[i] == ' ':
i += 1
# only contains blank space
if i == n:
return 0
sign = -1 if s[i] == '-' else 1
if s[i] in ['-', '+']:
i += 1
res, flag = 0, (2 ** 31 - 1) // 10
while i < n:
# not a number, exit the loop
if not s[i].isdigit():
break
c = int(s[i])
# if overflows
if res > flag or (res == flag and c > 7):
return 2 ** 31 - 1 if sign > 0 else -2 ** 31
res = res * 10 + c
i += 1
return sign * res
class Solution {
public int myAtoi(String s) {
if (s == null) return 0;
int n = s.length();
if (n == 0) return 0;
int i = 0;
while (s.charAt(i) == ' ') {
// only contains blank space
if (++i == n) return 0;
}
int sign = 1;
if (s.charAt(i) == '-') sign = -1;
if (s.charAt(i) == '-' || s.charAt(i) == '+') ++i;
int res = 0, flag = Integer.MAX_VALUE / 10;
for (; i < n; ++i) {
// not a number, exit the loop
if (s.charAt(i) < '0' || s.charAt(i) > '9') break;
// if overflows
if (res > flag || (res == flag && s.charAt(i) > '7')) return sign > 0 ? Integer.MAX_VALUE : Integer.MIN_VALUE;
res = res * 10 + (s.charAt(i) - '0');
}
return sign * res;
}
}
此处可能存在不合适展示的内容,页面不予展示。您可通过相关编辑功能自查并修改。
如您确认内容无涉及 不当用语 / 纯广告导流 / 暴力 / 低俗色情 / 侵权 / 盗版 / 虚假 / 无价值内容或违法国家有关法律法规的内容,可点击提交进行申诉,我们将尽快为您处理。