LeetCode——8.字符串转整数(atoi)【有穷状态自动机DFA】

题解

---

AC-Code

#include <bits/stdc++.h>
#include <unordered_map>
using namespace std;

typedef long long ll;

class Solution {
public:
	int myAtoi(string str) {
		Automation automation;
		for (char c : str)
			automation.get(c);
		return automation.ans * automation.sign;
	}

private:
	class Automation {
	private:
		string state = "start";

		unordered_map<string, vector<string>> table = {
			{"start",		{"start",	"signed",	"in_number",	"end"}},
			{"signed",		{"end",		"end",		"in_number",	"end"}},
			{"in_number",	{"end",		"end",		"in_number",	"end"}},
			{"end",			{"end",		"end",		"end",			"end"}}
		};

		int get_col(char c) {
			if (isspace(c))	return 0;
			if (c == '+' || c == '-')	return 1;
			if (isdigit(c))	return 2;
			return 3;
		}
	public:
		int sign = 1;
		ll ans = 0;

		void get(char c) {
			state = table[state][get_col(c)];
			if (state == "in_number") {
				ans = ans * 10 + c - '0';
				ans = sign == 1 ?
					min(ans, (ll)INT_MAX) :
					min(ans, -(ll)INT_MIN);
			}
			else if (state == "signed") {
				sign = c == '+' ? 1 : -1;
			}
		}
	};
};