标签：leetcode   算法   

问题描述：

Implement atoi to convert a string to an integer.

Hint: Carefully consider all possible input cases. If you want a challenge, please do not see below and ask yourself what are the possible input cases.

Notes: It is intended for this problem to be specified vaguely (ie, no given input specs). You are responsible to gather all the input requirements up front.

思路：

实现字符串转换为int型整数并不难，只是有好多中输入情况需要考虑，容易造成遗漏。首先看一下c++ 中atoi的定义：

-----------------------------------

atoi

int atoi (const char * str);

Return Value

------------------------------------------

可以看出需要注意的有以下几点：

1. 去掉前面的空格
2. 注意开始时候的符号 + -
3. 忽略数字后面的杂七杂八字符
4. 考虑溢出的情况。
   

考虑了上面几方面，代码就可以AC了。

代码：

public class String_To_Integer { //java
	public static int atoi(String str) {
		int haveMinus = 1;
		long result = 0;  
		
		if(str == null || str.trim().isEmpty())
			return 0;
		
		//chech + - 
		str = str.trim();
		if(str.charAt(0) == '-'){
			str = str.substring(1);
			haveMinus = -1;
		}else if(str.charAt(0) == '+')
			str = str.substring(1);
		
		//check num
		for(int i = 0;i < str.length() && str.charAt(i) >= '0' && str.charAt(i) <= '9'; i++){
			result = result*10 + (str.charAt(i)- '0');
		}
		
		//deal overflow
		if(result > 2147483647 && haveMinus == 1)
			return 2147483647;
		if(result > 2147483647 && haveMinus == -1)
			return -2147483648;
		
		return haveMinus*(int)result;
    }
	
	public static void main(String [] args){
		System.out.println(String_To_Integer.atoi("2147483648"));
	}
}

[leetcode]String to Integer (atoi)

标签：leetcode   算法   

原文地址：http://blog.csdn.net/chenlei0630/article/details/40714065