Algorithm

[LeetCode 58] Length of Last Word [Java]

顿搜

2018-01-29

0 评论

774 阅读

正在检测是否收录...

01/29

1. Description

Given a string s consists of upper/lower-case alphabets and empty space characters ' ', return the length of last word in the string.

If the last word does not exist, return 0.

Note: A word is defined as a character sequence consists of non-space characters only.

2. Example

Input: "Hello World"
Output: 5

3. Code


public class LeetCode0058 {
    public int lengthOfLastWord(String s) {
        if (s.length() == 0) {
            return 0;
        }

        boolean hasWord = false;
        int result = 0;

        for (int i = s.length() - 1; i >= 0; i--) {
            if (!hasWord) {
                if (s.charAt(i) == ' ') {
                    continue;
                } else {
                    hasWord = true;
                    result++;
                    continue;
                }
            }
            if (hasWord) {
                if (s.charAt(i) != ' ') {
                    result++;
                } else {
                    break;
                }
            }
        }
        return result;
    }

    public static void main(String[] args) {
        LeetCode0058 leetcode = new LeetCode0058();
        System.out.println(leetcode.lengthOfLastWord("Hello World"));
    }
}

String

朗读