434. Number of Segments in a String

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Description

Given a string s, return the number of segments in the string.

A segment is defined to be a contiguous sequence of non-space characters.

 

Example 1:

Input: s = "Hello, my name is John"
Output: 5
Explanation: The five segments are ["Hello,", "my", "name", "is", "John"]

Example 2:

Input: s = "Hello"
Output: 1

 

Constraints:

• 0 <= s.length <= 300
• s consists of lowercase and uppercase English letters, digits, or one of the following characters "!@#$%^&*()_+-=',.:".
• The only space character in s is ' '.

Solutions

Solution 1: String Splitting

We split the string $\textit{s}$ by spaces and then count the number of non-empty words.

The time complexity is $O(n)$, and the space complexity is $O(n)$, where $n$ is the length of the string $\textit{s}$.

Python3

class Solution:
    def countSegments(self, s: str) -> int:
        return len(s.split())

Java

class Solution {
    public int countSegments(String s) {
        int ans = 0;
        for (String t : s.split(" ")) {
            if (!"".equals(t)) {
                ++ans;
            }
        }
        return ans;
    }
}

C++

class Solution {
public:
    int countSegments(string s) {
        int ans = 0;
        istringstream ss(s);
        while (ss >> s) ++ans;
        return ans;
    }
};

Go

func countSegments(s string) int {
	ans := 0
	for _, t := range strings.Split(s, " ") {
		if len(t) > 0 {
			ans++
		}
	}
	return ans
}

TypeScript

function countSegments(s: string): number {
    return s.split(/\s+/).filter(Boolean).length;
}

PHP

class Solution {
    /**
     * @param String $s
     * @return Integer
     */
    function countSegments($s) {
        $arr = explode(' ', $s);
        $cnt = 0;
        for ($i = 0; $i < count($arr); $i++) {
            if (strlen($arr[$i]) != 0) {
                $cnt++;
            }
        }
        return $cnt;
    }
}

Solution 2: Simulation

We can also directly traverse each character $\text{s[i]}$ in the string. If $\text{s[i]}$ is not a space and $\text{s[i-1]}$ is a space or $i = 0$, then $\text{s[i]}$ marks the beginning of a new word, and we increment the answer by one.

After the traversal, we return the answer.

The time complexity is $O(n)$, where $n$ is the length of the string $\textit{s}$. The space complexity is $O(1)$.

Python3

class Solution:
    def countSegments(self, s: str) -> int:
        ans = 0
        for i, c in enumerate(s):
            if c != ' ' and (i == 0 or s[i - 1] == ' '):
                ans += 1
        return ans

Java

class Solution {
    public int countSegments(String s) {
        int ans = 0;
        for (int i = 0; i < s.length(); ++i) {
            if (s.charAt(i) != ' ' && (i == 0 || s.charAt(i - 1) == ' ')) {
                ++ans;
            }
        }
        return ans;
    }
}

C++

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

Go

func countSegments(s string) int {
	ans := 0
	for i, c := range s {
		if c != ' ' && (i == 0 || s[i-1] == ' ') {
			ans++
		}
	}
	return ans
}

TypeScript

function countSegments(s: string): number {
    let ans = 0;
    for (let i = 0; i < s.length; i++) {
        let c = s[i];
        if (c !== ' ' && (i === 0 || s[i - 1] === ' ')) {
            ans++;
        }
    }
    return ans;
}

PHP

class Solution {
    /**
     * @param String $s
     * @return Integer
     */
    function countSegments($s) {
        $ans = 0;
        $n = strlen($s);
        for ($i = 0; $i < $n; $i++) {
            $c = $s[$i];
            if ($c !== ' ' && ($i === 0 || $s[$i - 1] === ' ')) {
                $ans++;
            }
        }
        return $ans;
    }
}