Leetcode Valid Number

题目

 

思路

1. WA 了十次

2. 

// e9 False
// .1 true
// 3. true
// . false
// 343e+34 true

3. 分成了三个部分, 三部分之间有较弱的依赖关系, 还可以通过 bool 变量与返回值分开

 

代码

bool pa, pb, pc;
int FurtherA(const char *s, int &st, int ed)  {
	if(s[st] == '+' || s[st] == '-')
		st ++;

	if(st > ed) return -1;

	if(s[st] == '.')  {
		st ++;
		return 1;
	}  else  {
		if(!(s[st] <= '9' && s[st] >= '0')) return -1;
	}

	
	while(st <= ed)  {
		if(s[st] >= '0' && s[st] <= '9')  {
			st ++;
			pa = true;
			continue;
		}
		if(s[st] == '.')  {
			st ++;
			return 1;
		}
		if(s[st] == 'e')  {
			st ++;
			return 2;
		}
		return -1;
	}

	return 0;

}

int FurtherB(const char *s, int &st, int ed)  {
	if(st > ed) return 0;
	//if(!(s[st] <= '9' && s[st] >= '0')) return -1;

	while(st <= ed)  {
		if(s[st] >= '0' && s[st] <= '9')  {
			st ++;
			pb = true;
			continue;
		}

		if(s[st] == 'e')  {
			st ++;
			return 2;
		}
		// else
		return -1;
	}
		return 0;
}

int FurtherC(const char *s, int &st, int ed)  {
	if(s[st] == '+' || s[st] == '-') 
		st ++;
	
	if(st > ed) return -1;
	if(!(s[st] <= '9' && s[st] >= '0')) return -1;
	while(st <= ed)  {
		if(s[st] >= '0' && s[st] <= '9') {
			st ++;
			pc = true;
			continue;
		}
		return -1;
	}

	return 0;
}


class Solution {
public:
    bool isNumber(const char *s) {
		int len = strlen(s);

		int i = 0, j = len-1;
		// skip 
		while(i < len && s[i] == ' ')  {
			i ++;
		}        
		while(j >= i && s[j] == ' ')  {
			j --;
		}

		if(i > j) return false;

		if(i == j && s[i] == '.')
			return false;

		pa = pb = pc = false;

		int res_a, res_b = 1, res_c = 1;

		res_a = FurtherA(s, i, j);

		//cout << res_a << endl;
		if(res_a == 0 && pa)  {
			return true;
		}  else if  (res_a == -1)  {
			return false;
		}  else if  (res_a == 1)  {
			res_b = FurtherB(s, i, j);
		}

		if(res_b == 0 && (pa || pb)) {
			return true;
		}  else if(res_b == -1)  {
			return false;
		}

		if(res_a == 2 || res_b == 2)  {
			res_c = FurtherC(s, i, j);
		}

		if(res_c == 0 && pc && (pa || pb ))  {
			return true;
		}

		return false;
    }
};

  

posted @ 2014-04-23 23:59  周卓  阅读(131)  评论(0编辑  收藏  举报