[LeetCode] Length of Last Word 字符串查找

Given a string s consists of upper/lower-case alphabets and empty space characters ' ', return the length of last word in the string.

If the last word does not exist, return 0.

Note: A word is defined as a character sequence consists of non-space characters only.

For example, 
Given s = "Hello World",
return 5.

Hide Tags

 String

 

---

    这题比较简单，只是可能前面有空格，后面有空格。- -

算法逻辑：

1. 排除最后的空格
2. index 从后往前查第一个空格。
3. 返回长度。

#include <iostream>
#include <cstring>
using namespace std;


class Solution {
public:
    int lengthOfLastWord(const char *s) {
        int n = strlen(s);
        if(n <1)    return 0;
        int idx=n-1;
        while(idx>=0&&s[idx]==' ')  idx--;
        n = idx+1;
        while(idx>=0){
            if (s[idx]==' ')  break;
            idx --;
        }
//        if(idx<0)   return 0;
        return n - idx -1;
    }
};

int main()
{
    char s[] = " 12   ";
    Solution sol;
    cout<<sol.lengthOfLastWord(s)<<endl;
    return 0;
}