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                合規國際互聯網加速 OSASE為企業客戶提供高速穩定SD-WAN國際加速解決方案。 廣告
                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`. ~~~ class Solution { public: int lengthOfLastWord(const char *s) { int ans = 0, prev = 0; bool space = false; for(int i = 0; s[i] != NULL; ++i){ if(s[i] == ' ') { if(space == false)prev = ans, ans = 0, space = true; else continue; }else ans++, space = false; } if(ans == 0) return prev; return ans; } }; ~~~ ? ?
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                              哎呀哎呀视频在线观看