C++实现LeetCode(58.求末尾单词的长度)

[LeetCode] 58. Length of Last Word 求末尾单词的长度

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.

Example:

Input: "Hello World"
Output: 5

这道题难度不是很大。先对输入字符串做预处理，去掉开头和结尾的空格，然后用一个计数器来累计非空格的字符串的长度，遇到空格则将计数器清零，参见代码如下：

解法一：

class Solution {
public:
    int lengthOfLastWord(string s) {
        int left = 0, right = (int)s.size() - 1, res = 0;
        while (s[left] == ' ') ++left;
        while (s[right] == ' ') --right;
        for (int i = left; i <= right; ++i) {
            if (s[i] == ' ') res = 0;
            else ++res;
        }
        return res;
    }
};

昨晚睡觉前又想到了一种解法，其实不用上面那么复杂的，这里关心的主要是非空格的字符，那么实际上在遍历字符串的时候，如果遇到非空格的字符，只需要判断其前面一个位置的字符是否为空格，如果是的话，那么当前肯定是一个新词的开始，将计数器重置为1，如果不是的话，说明正在统计一个词的长度，计数器自增1即可。但是需要注意的是，当 i=0 的时候，无法访问前一个字符，所以这种情况要特别判断一下，归为计数器自增1那类，参见代码如下：

解法二：

class Solution {
public:
    int lengthOfLastWord(string s) {
        int res = 0;
        for (int i = 0; i < s.size(); ++i) {
            if (s[i] != ' ') {
                if (i != 0 && s[i - 1] == ' ') res = 1;
                else ++res;
            }
        }
        return res;
    }
};

下面这种方法是第一种解法的优化版本，由于只关于最后一个单词的长度，所以开头有多少个空格起始并不需要在意，从字符串末尾开始，先将末尾的空格都去掉，然后开始找非空格的字符的长度即可，参见代码如下：

解法三：

class Solution {
public:
    int lengthOfLastWord(string s) {
        int right = s.size() - 1, res = 0;
        while (right >= 0 && s[right] == ' ') --right;
        while (right >= 0 && s[right] != ' ' ) {
            --right;
            ++res;
        }
        return res;
    }
};

这道题用Java来做可以一行搞定，请参见这个帖子.

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