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Commit d591cf6

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Merge pull request #1 from borrelunde/string_to_integer_atoi
LeetCode problem: 8. String to Integer (atoi)
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# 8. String to Integer (atoi)
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Difficulty: `Medium`
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Topics: `String`
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Implement the `myAtoi(string s)` function, which converts a string to a 32-bit signed integer (similar to C/C++'s `atoi`
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function).
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The algorithm for `myAtoi(string s)` is as follows:
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1. Read in and ignore any leading whitespace.
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2. Check if the next character (if not already at the end of the string) is `'-'` or `'+'`. Read this character in if it
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is either. This determines if the final result is negative or positive respectively. Assume the result is positive if
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neither is present.
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3. Read in next the characters until the next non-digit character or the end of the input is reached. The rest of the
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string is ignored.
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4. Convert these digits into an integer (i.e. `"123" -> 123`, `"0032" -> 32`). If no digits were read, then the integer
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is `0`. Change the sign as necessary (from step 2).
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5. If the integer is out of the 32-bit signed integer range `[-2^31, 2^31 - 1]`, then clamp the integer so that it
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remains
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in the range. Specifically, integers less than `-2^31` should be clamped to `-2^31`, and integers greater
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than `2^31 - 1` should be clamped to `2^31 - 1`.
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6. Return the integer as the final result.
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Note:
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- Only the space character `' '` is considered a whitespace character.
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- **Do not ignore** any characters other than the leading whitespace or the rest of the string after the digits.
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**Example 1:**
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```text
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Input: s = "42"
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Output: 42
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Explanation: The caret is the current reader position.
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Step 1: "42" (no characters read because there is no leading whitespace)
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^
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Step 2: "42" (no characters read because there is neither a '-' nor '+')
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^
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Step 3: "42" ("42" is read in)
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^
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The parsed integer is 42.
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Since 42 is in the range [-2^31, 2^31 - 1], the final result is 42.
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```
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**Example 2:**
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```text
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Input: s = " -42"
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Output: -42
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Explanation:
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Step 1: " -42" (leading whitespace is read and ignored)
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^
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Step 2: " -42" ('-' is read, so the result should be negative)
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^
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Step 3: " -42" ("42" is read in)
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^
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The parsed integer is -42.
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Since -42 is in the range [-2^31, 2^31 - 1], the final result is -42.
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```
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**Example 3:**
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```text
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Input: s = "4193 with words"
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Output: 4193
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Explanation:
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Step 1: "4193 with words" (no characters read because there is no leading whitespace)
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^
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Step 2: "4193 with words" (no characters read because there is neither a '-' nor '+')
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^
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Step 3: "4193 with words" ("4193" is read in; reading stops because the next character is a non-digit)
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^
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The parsed integer is 4193.
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Since 4193 is in the range [-2^31, 2^31 - 1], the final result is 4193.
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```
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**Constraints:**
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- `0 <= s.length <= 200`
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- `s` consists of English letters (lower-case and upper-case), digits (`0-9`), `' '`, `'+'`, `'-'`, and `'.'`.
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package com.bl.string_to_integer_atoi;
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/**
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* This is the solution to the LeetCode problem: 8. String to Integer (atoi)
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*
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* @author Børre A. Opedal Lunde
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* @since 2024年01月25日
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*/
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public class Solution {
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static final char WHITESPACE = ' ';
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static final char PLUS = '+';
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static final char MINUS = '-';
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static final int MINIMUM_VALUE = 0x80000000; // -2^31
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static final int MAXIMUM_VALUE = 0x7fffffff; // 2^31 - 1
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public int myAtoi(String s) {
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// If the string is empty, return zero.
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if (s.isEmpty()) {
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return 0;
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}
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// Convert the string to a character array that we'll iterate over.
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final char[] characters = s.toCharArray();
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// Used to keep track of whether the number is negative or not.
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boolean negative = false;
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// This is the index used to iterate over the character array. It is
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// declared outside the loop so that it can be used in both loops.
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int i = 0;
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// First, ignore leading whitespace and account for explicit sign
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// symbol.
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for (; i < characters.length; i++) {
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char character = characters[i];
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// Ignore leading whitespace.
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if (character == WHITESPACE) {
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continue;
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}
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// Account for explicit sign symbol.
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else if (character == PLUS) {
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// The number is positive by default, so we don't need to do
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// set the negative flag to false.
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// Increment the index to skip the sign symbol.
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i++;
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} else if (character == MINUS) {
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negative = true;
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// Same as above.
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i++;
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}
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// If the character is not whitespace nor a sign symbol, we will
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// delegate the rest of the work to the next loop. It is responsible
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// for parsing the number.
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break;
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}
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// We're parsing an integer, but we're using a long to avoid overflow
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// and underflow.
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long number = 0;
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// Parse the number.
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for (; i < characters.length; i++) {
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final char character = characters[i];
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// If the character is not a digit, we will break out of the loop.
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if (! characterIsDigit(character)) {
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break;
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}
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// Convert the character to a digit.
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final int digit = digitFromCharacter(character);
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// First, multiply the number by ten to make room for the new digit.
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// Then, add the digit to the number, taking into account whether
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// the number is negative or not.
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number = number * 10 + (negative ? - digit : digit);
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// Clamp the number if it is greater than the maximum value.
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if (number <= MINIMUM_VALUE) {
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return MINIMUM_VALUE;
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}
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// Clamp the number if it is less than the minimum value.
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else if (number >= MAXIMUM_VALUE) {
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return MAXIMUM_VALUE;
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}
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}
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// Return the number as an integer.
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return (int) number;
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}
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static int digitFromCharacter(final char character) {
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// Assumes the digit character is in the ASCII table.
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return character - '0';
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}
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static boolean characterIsDigit(final char character) {
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// Assumes the character is in the ASCII table.
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return character >= '0' && character <= '9';
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}
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}

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