Daily practice (38): The length of the last word

---

title: Daily practice (38): the length of the last word

categories:[Swords offer]

tags:[Daily practice]

date: 2022/04/09

---

Daily practice (38): The length of the last word

You are given a string s consisting of several words separated by some space characters before and after them. Returns the length of the last word in the string.

A word is the largest substring consisting of only letters and not containing any space characters.

Example 1:

Input: s = "Hello World"

output: 5

Explanation: The last word is "World" and has a length of 5.

Example 2:

Input: s = "fly me to the moon"

output: 4

Explanation: The last word is "moon" and has a length of 4.

Example 3:

Input: s = "luffy is still joyboy"

Output: 6

Explanation: The last word is "joyboy" of length 6.

hint:

1 <= s.length <= 104

s only consists of English letters and spaces ' '

there is at least one word in s

Source: LeetCode

Link: https://leetcode-cn.com/problems/length-of-last-word

Method 1: Double pointer method

Thought analysis

Define two pointers to point to the end of the string, if the end of the string is a space, the pointers move forward at the same time i--; j--;

If a non-space is encountered, then the fixed j does not move, indicating the end of the word, and continues to move forward i until a space is encountered again, or the string is traversed;

Finally, the position pointed to by i is the position before the beginning of the last word, and j - i is the length of the last word.

 int lengthOfLastWord(string s) {
    int i = s.size() - 1;    //定义两个指针指向字符串的末尾
    int j = i;
    while (s[i] == ' ') {
        i--;
        j--;
    }
    //此时i,j指向最后一个单词的末尾位置
    while (i >= 0 && s[i] != ' ') {
        i--;
    }
    return j - i; //得到最后一个单词的长度
}

Method 2: Reverse Traversal

Thought analysis

Continue traversing the string in reverse, starting at the last letter, until a space is encountered or the beginning of the string is reached. Each letter traversed is a letter in the last word, so the number of letters traversed is

the length of the last word

 int lengthOfLastWord(string s) {
    int index = s.size() - 1;
    while (s[index] == ' ') {
        index--;
    }
    //此时index指向最后一个字符串的的末尾
    int wordLength = 0;
    while (index >= 0 && s[index] != ' ') {
        wordLength++;    //通过遍历得到最后一个字符串的长度
        index--;
    }
    return wordLength; 
}