【leetcode】58. Length of Last Word(Python & C++)
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58. Length of Last Word
题目链接
58.1 题目描述:
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.
58.2 解题思路:
思路一:从后往前遍历。首先找到字符串中第一不为’ ‘的值的坐标,该坐标即为最后一个Word的尾字母。然后继续遍历,如果碰到’ ‘,则遍历结束,否则,count++。最后返回count。
思路二:同思路一相同,不过写法上更优雅。设置count=0,first=s.length() - 1。第一个遍历从后向前遍历,如果first大于等于0,且其字符等于’ ‘,则first–;第二个遍历,如果first大于等于0,且其字符不等于’ ‘,则first–,count++。最后返回count。
58.3 C++代码:
1、思路一代码(6ms):
class Solution115 {public: int lengthOfLastWord(string s) { if (s.length() == 0) return 0; int first = -1; for (int i = s.length()-1; i >= 0; i--) { if (s[i] != ' ') { first = i; break; } } if (first == -1) return 0; int count = 0; for (int i = first; i >=0 ; i--) { if (s[i] == ' ') break; else count++; } return count; }};
2、思路二代码(6ms)
class Solution115_1 {public: int lengthOfLastWord(string s) { int count = 0; int first = s.length() - 1; while (first >= 0 && s[first] == ' ') first--; while (first >= 0 && s[first] != ' ') { count++; first--; } return count; }};
58.4 Python代码:
2、思路二代码(35ms)
class Solution(object): def lengthOfLastWord(self, s): """ :type s: str :rtype: int """ count=0 first=len(s)-1 while first>=0 and s[first]==' ': first-=1 while first>=0 and s[first]!=' ': first-=1 count+=1 return count
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