标签：exist note lower while har ase sts rac 循环
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",
return5.
解法一：
顺序遍历，有单词则len++，有空格且下一位为单词则len=0，直到末尾输出len
class Solution { public: int lengthOfLastWord(const char *s) { int len=0; while(*s) { if(*s++!=‘ ‘) len++; else if(*s&&*s!=‘ ‘) len=0; } return len; } };
解法二：
class Solution { public: int lengthOfLastWord(const char *s) { int count = 0; int len = strlen(s); //反向查找，末尾空格忽略，行中出现空格就终止循环 for(int i = len-1; i >= 0 ; i--){ if(s[i] == ‘ ‘){ if(count) break; } else{ count++; } } return count; } };
标签：exist note lower while har ase sts rac 循环
原文地址：https://www.cnblogs.com/zl1991/p/9638236.html
