LeetCode --- 58. Length of Last Word
2015-03-14 23:13
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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,
这道题的要求是返回字符串中最后一个单词的长度。
遍历字符串,统计单词长度。只有一点需要注意,就是最后单词后面有空格的情况,因此需要判断下一字符是否是字母,如果是,这重置长度为0即可。
时间复杂度:O(n)
空间复杂度:O(1)
当然,也可以引入计数器遍历cnt用于记录当前单词长度,只有当cnt加1的时候,更新结果变量res。
转载请说明出处:LeetCode --- 58. Insert Interval
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,
[code]Given s = "Hello World", return 5.
这道题的要求是返回字符串中最后一个单词的长度。
遍历字符串,统计单词长度。只有一点需要注意,就是最后单词后面有空格的情况,因此需要判断下一字符是否是字母,如果是,这重置长度为0即可。
时间复杂度:O(n)
空间复杂度:O(1)
[code] 1 class Solution 2 { 3 public: 4 int lengthOfLastWord(const char *s) 5 { 6 int l = 0; 7 while(*s) 8 { 9 if(' ' != *s ++) 10 ++ l; 11 else if(*s && ' ' != *s) 12 l = 0; 13 } 14 return l; 15 } 16 };
当然,也可以引入计数器遍历cnt用于记录当前单词长度,只有当cnt加1的时候,更新结果变量res。
[code] 1 class Solution 2 { 3 public: 4 int lengthOfLastWord(const char *s) 5 { 6 int res = 0, cnt = 0; 7 while(*s) 8 { 9 if(' ' == *s ++) 10 cnt = 0; 11 else 12 res = ++ cnt; 13 } 14 return res; 15 } 16 };
转载请说明出处:LeetCode --- 58. Insert Interval
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