58. Length of Last Word

Problem

View on LeetCode →

Difficulty: Easy
Tags: String

Intuition

the first intuition to this problem is that I should use variable that stores previous character. because new word starts not with a non-space letter, but a non-space letter after a space. with this idea, I could solve this very easily.

Approach

As I described on the intuition part, i made a variable called ‘prev’, which stores previous character. the initial value of it should be a space, so that s starting with non-space letter should be recognized as a word. other parts are very straight forward so esay.

Solution

int lengthOfLastWord(char* s) {
    char prev = ' ';
    int i = 0;
    int ans = 0;
    while(s[i] != '\0'){
        if(s[i] != ' '){
            if(prev == ' '){
                ans = 0;
            }
            ans++;
        }
        prev = s[i];
        i++;
    }
    return ans;
}

Complexity

• Time: $O(n)$ where n is length of s

• Space: $O(1)$

Thoughts

yup. this was easier than 011, but worth problem I think.