LeetCode-434-Number of words in a string

The number of words in the string

Title description: Count the number of words in a string, where words refer to consecutive characters that are not spaces.

Please note that you can assume that the string does not contain any non-printable characters.

Please refer to LeetCode official website for example description.

Source: LeetCode
Link: https://leetcode-cn.com/problems/number-of-segments-in-a-string/
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Solution one: string traversal

First, if s is null or s is an empty string, then 0 is directly returned.

Otherwise, a statement COUNT recording the number of words is initialized to 0, lastChar recording the initial value of a character space character , then traverse S characters C , process is as follows:

• If c and lastChar are both spaces, it is currently impossible to be a word, skip;
• If the previous character is a space and the current character is not a space, the current character is the beginning of a word, count plus one, and the lastChar is updated to the current character;
• If neither the previous character nor the current character is a space, skip it;
• If the previous character is not a space and the current character is a space, the previous character is the last character of the previous word. lastChar to the current character.

Finally, return count is the number of words in the string s.

/**
 * @Author: ck
 * @Date: 2021/9/29 8:51 下午
 */
public class LeetCode_434 {
    public static int countSegments(String s) {
        if (s == null || s.length() == 0) {
            return 0;
        }
        int count = 0;
        char lastChar = ' ';
        for (char c : s.toCharArray()) {
            if (lastChar == ' ' && c == ' ') {
                // 如果上一个字符和当前字符都是空格，则跳过
                continue;
            } else if (lastChar == ' ' && c != ' ') {
                // 如果上一个字符是空格，当前字符不是空格，则当前字符是一个单词的开始，count加一，并且将lastChar更新为当前字符
                lastChar = c;
                count++;
            } else if (lastChar != ' ' && c != ' ') {
                // 如果上一个字符和当前字符都不是空格，则跳过
                continue;
            } else if (lastChar != ' ' && c == ' ') {
                // 如果上一个字符不是空格，而当前字符是空格，则上一个字符是上一个单词的最后一个字符。将lastChar更新为当前字符
                lastChar = c;
            }
        }
        return count;
    }

    public static void main(String[] args) {
        // 期望输出: 5
        System.out.println(countSegments("Of all the gin joints in all the towns in all the world,  "));
    }
}

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