[LeetCode]58.Length of Last Word

題目

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.

分析

無

代碼

    /**------------------------------------
    *   日期：2015-02-06
    *   作者：SJF0115
    *   題目: 58.Length of Last Word
    *   網址：https://oj.leetcode.com/problems/length-of-last-word/
    *   結果：AC
    *   來源：LeetCode
    *   博客：
    ---------------------------------------**/
    #include 
    #include 
    #include 
    #include 
    using namespace std;

    class Solution {
    public:
        int lengthOfLastWord(const char *s) {
            int len = strlen(s);
            int i = len - 1;
            int lastLen = 0;
            // 去掉空格
            while(s[i] == ' '){
                --i;
            }//while
            while(i >= 0 && s[i] != ' '){
                ++lastLen;
                --i;
            }//while
            return lastLen;
        }
    };

    int main(){
        Solution s;
        char *str = "    q  f  ";//"hello world";
        int result = s.lengthOfLastWord(str);
        // 輸出
        cout<