Length of Last Word
本题收获：
1.str.size()为负
2.size_t引发的死循环
3.题目思路，有时在写代码时很不清楚边界条件的输出值是什么，若为面试，一定要问清楚。
题目：
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.
边界条件：
1. " " return 0
2."a " return 1 (我开始错以为这种情况返回0 )
思路：
我的思路：直接找，for循环
leetcode：直接找，while循环
正确代码：
1 class Solution { 2 public: 3 int lengthOfLastWord(string s) { 4 int len = 0, tail = s.length() - 1; 5 while (tail >= 0 && s[tail] == ' ') tail--; 6 while (tail >= 0 && s[tail] != ' ') { 7 len++; 8 tail--; 9 } 10 return len; 11 } 12 };
我写的：
1 class MyClass 2 { 3 public: 4 int lengthOfLastWord(string str) 5 { 6 if (str.size() == 0) return 0; 7 int n = str.size() - 1; 8 int j = 0; 9 10 while (n >= 0 && str[n] == ' ') 11 { 12 n--; 13 } 14 while (n >= 0 && str[n] != ' ') 15 { 16 j++; 17 n--; 18 } 19 return j; 20 } 21 };
出现死循环的代码：
1 class MyClass 2 { 3 public: 4 int lengthOfLastWord(string str) 5 { 6 if (str.size() == 0) return 0; 7 int j = 0; 8 int n = str.size() - 1; 9 for (size_t i = n; i >= 0; i--) //从size_t换到 int就不会出现i = 4294967295 10 { 11 cout << i << endl; 12 if (str[n] == ' ') return 0; 13 else 14 { 15 j++; 16 continue; 17 } 18 19 if (str[i] == ' ') break; 20 else j++; 21 } 22 return j; 23 } 24 };
因为size_t是unsigned int/unsigned long 型 ，好吧 具体我也不太清楚是怎么样的！
测试全代码：
1 #include "stdafx.h" 2 #include "iostream" 3 #include "string" 4 using namespace std; 5 6 class MyClass 7 { 8 public: 9 int lengthOfLastWord(string str) 10 { 11 if (str.size() == 0) return 0; 12 int n = str.size() - 1; 13 int j = 0; 14 15 while (n >= 0 && str[n] == ' ') 16 { 17 n--; 18 } 19 while (n >= 0 && str[n] != ' ') 20 { 21 j++; 22 n--; 23 } 24 return j; 25 } 26 }; 27 28 int _tmain(int argc, _TCHAR* argv[]) 29 { 30 string str; 31 str = "a"; 32 MyClass solution; 33 int m = 0; 34 m = solution.lengthOfLastWord(str); 35 cout << m << endl; 36 system("pause"); 37 return 0; 38 }