闯关leetcode——58. Length of Last Word

大纲

• 题目
• • 地址
  • 内容
• 解题
• • 代码地址

题目

地址

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

内容

Given a string s consisting of words and spaces, return the length of the last word in the string.

A word is a maximal substring consisting of non-space characters only.

Example 1:

Input: s = “Hello World”
Output: 5
Explanation: The last word is “World” with length 5.

Example 2:

Input: s = " fly me to the moon "
Output: 4
Explanation: The last word is “moon” with length 4.

Example 3:

Input: s = “luffy is still joyboy”
Output: 6
Explanation: The last word is “joyboy” with length 6.

Constraints:

• 1 <= s.length <= 10⁴
• s consists of only English letters and spaces ’ '.
• There will be at least one word in s.

解题

这题就是要找到一个字符串中最后一段非空格的连续字符长度。我们只要定义好这样的子串特点即可：以空格后的第一个字符开始，以空格前的最后一个字符结束。向后遍历，找到最后一个符合这个特点的子串。需要注意的是两个边界：我们可以假设原始字符串前有一个空格，原始字符串后有一个空格，这可以保证我们逻辑的一致性。

#include <string>
using namespace std;

class Solution {
public:
    int lengthOfLastWord(string s) {
        bool preIsSpace = true; // if the previous character is space
        int cur = 0;
        int lastWordStart = 0;
        int lastWordEnd = 0;
        for (char c : s) {
            if (c == ' ') {
                if (!preIsSpace) {
                    lastWordEnd = cur;
                }
                preIsSpace = true;
            } else {
                if (preIsSpace) {
                    lastWordStart = cur;
                }
                preIsSpace = false;
            }
            cur++;
        }
        if (!preIsSpace) {
            lastWordEnd = cur;
        }
        return lastWordEnd - lastWordStart;
    }
};

代码地址

https://github.com/f304646673/leetcode/tree/main/58-Length-of-Last-Word