【LeetCode】(58)Length of Last Word(Easy)
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题目
Length of Last Word
Total Accepted: 59959 Total Submissions: 218191My SubmissionsGiven 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
.
解析
只需要用一个i从后往前找,跳过空格,找到字符,然后遇到空格停止即可,需要注意边界条件。题目很简单,代码如下
class Solution {public: int lengthOfLastWord(string s) { int i = s.size()-1;int ret = 0;while ( i>=0 && s[i] == ' ') i--; while (i>=0 && s[i]!=' ') { i--; ret++; } return ret; }};
看看大神代码
public:int lengthOfLastWord(const char *s) { const string str(s); auto first = find_if(str.rbegin(), str.rend(), ::isalpha); auto last = find_if_not(first, str.rend(), ::isalpha); return distance(first, last); }};和
class Solution {public:int lengthOfLastWord(const char *s) { int len = 0; while (*s) { if (*s++ != ' ') ++len; else if (*s && *s != ' ') len = 0; } return len;}};
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