【LeetCode】Length of Last Word

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Length of Last Word

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.

 

用result記錄上一個單詞的長度，用cur記錄當前單詞的長度。

每遇到空白符，將cur賦給result，然後cur置零。

如果以字母結尾，返回cur

如果以空格結尾，返回result(此時cur為0)

 

class Solution {public:    int lengthOfLastWord(const char *s) {        int result = 0; //last word length        if(s == NULL)            return result;        int i = 0;        int cur = 0;    //cur word length        while(s[i] != 0)        {            //skip space            while(s[i] != 0 && s[i] == ‘ ‘)                i ++;            if(s[i] == 0)                return result;                            while(s[i] != 0 && s[i] != ‘ ‘)            {//cur word                i ++;                cur ++;            }            if(s[i] == 0)   //end            {                return cur; //cur is exactly the last word            }            else            {//space remains to be addressed in next loop                result = cur;                cur = 0;            }        }    }};

【LeetCode】Length of Last Word