标签：
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.
解法1：对输入字符串从后往前扫描，先去掉末尾空格得到输入字符串的“真正”end，然后找到第一个空格位置i，两个index之间即为最后一个单词，返回其长度即可。
class Solution { public: int lengthOfLastWord(string s) { int n = s.size(), end = n - 1; int i = n - 1, j = n - 2; if (n == 0) return 0; for (; i >= 0; --i) { if (i == n - 1 && s[i] == ‘ ‘) { while (s[j] == s[j + 1]) --j; if (j < 0) return 0; i -= end - j; end = j; } if (s[i] == ‘ ‘) return end - i; } return end - i; } };
解法2：另一个思路是先对输入字符串预处理，去掉开头和结尾多余的空格，然后从前往后扫描，遇到空格将计数器置零，否则计数器自增。这样计数器记录的自然就是最后一个单词的长度。
class Solution { public: int lengthOfLastWord(string s) { int n = s.size(), count = 0; int left = 0, right = n - 1; while (left < n && s[left] == ‘ ‘) ++left; while (right >= 0 && s[right] == ‘ ‘) --right; for (int i = left; i <= right; ++i) { if (s[i] != ‘ ‘) ++count; else count = 0; } return count; } };
[LeetCode]38. Length of Last Word最后单词长度
标签：
原文地址：http://www.cnblogs.com/aprilcheny/p/4907313.html
