8. 字符串转换整数 (atoi)
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字符串
中频
解法一:自动机
时间复杂度:O(n) | 空间复杂度:O(1) | 推荐使用
动画演示
准备就绪 - 输入字符串,然后点击开始
代码实现
class Solution {
public int myAtoi(String str) {
Automaton automaton = new Automaton();
int length = str.length();
for (int i = 0; i < length; ++i) {
automaton.get(str.charAt(i));
}
return (int) (automaton.sign * automaton.ans);
}
}
class Automaton {
public int sign = 1;
public long ans = 0;
private String state = "start";
private Map<String, String[]> table = new HashMap<String, String[]>() {{
put("start", new String[]{"start", "signed", "in_number", "end"});
put("signed", new String[]{"end", "end", "in_number", "end"});
put("in_number", new String[]{"end", "end", "in_number", "end"});
put("end", new String[]{"end", "end", "end", "end"});
}};
public void get(char c) {
state = table.get(state)[get_col(c)];
if ("in_number".equals(state)) {
ans = ans * 10 + c - '0';
ans = sign == 1 ? Math.min(ans, (long) Integer.MAX_VALUE) : Math.min(ans, -(long) Integer.MIN_VALUE);
} else if ("signed".equals(state)) {
sign = c == '+' ? 1 : -1;
}
}
private int get_col(char c) {
if (c == ' ') {
return 0;
}
if (c == '+' || c == '-') {
return 1;
}
if (Character.isDigit(c)) {
return 2;
}
return 3;
}
}
// 测试示例
public class Main {
public static void main(String[] args) {
Solution solution = new Solution();
System.out.println(solution.myAtoi("42")); // 输出: 42
System.out.println(solution.myAtoi(" -42")); // 输出: -42
System.out.println(solution.myAtoi("4193 with words")); // 输出: 4193
System.out.println(solution.myAtoi("words and 987")); // 输出: 0
System.out.println(solution.myAtoi("-91283472332")); // 输出: -2147483648
}
}时间复杂度:O(n)
空间复杂度:O(1)
本实现使用有限状态自动机来处理字符串到整数的转换。自动机包含四个状态:start(初始状态)、signed(处理符号)、in_number(处理数字)和end(结束状态)。通过状态转换表,根据当前状态和输入字符确定下一个状态,并在相应状态下执行不同的操作。