LeetCode: 58. Length of Last Word

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58. Length of Last Word

Given a string s consists of upper/lower-case alphabets and empty space characters ' ', return the length of last word (last word means the last appearing word if we loop from left to right) in the string.

If the last word does not exist, return 0.

Note: A word is defined as a maximal substring consisting of non-space characters only.

Example:

Input: "Hello World"
Output: 5

代码还是要力求简洁明了！

If the last word does not exist, return 0.

Note: A word is defined as a maximal substring consisting of non-space characters only.

Example:

Input: "Hello World"
Output: 5

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/**
 * Runtime: 0 ms, faster than 100.00% of Java online submissions for Length of Last Word.
 * Memory Usage: 38 MB, less than 7.57% of Java online submissions for Length of Last Word.
 *
 * Copy from: https://leetcode-cn.com/problems/length-of-last-word/solution/hua-jie-suan-fa-58-zui-hou-yi-ge-dan-ci-de-chang-d/[画解算法：58. 最后一个单词的长度 - 最后一个单词的长度 - 力扣（LeetCode）]
 */
public int lengthOfLastWord(String s) {
    if (Objects.isNull(s) || s.length() == 0) {
        return 0;
    }
    int end = s.length() - 1;
    while (end >= 0 && s.charAt(end) == ' ') {
        end--;
    }
    if (end < 0) {
        return 0;
    }
    int start = end;
    while (start >= 0 && s.charAt(start) != ' ') {
        start--;
    }
    return end - start;
}

/**
 * Runtime: 0 ms, faster than 100.00% of Java online submissions for Length of Last Word.
 * Memory Usage: 37.3 MB, less than 48.48% of Java online submissions for Length of Last Word.
 */
public int lengthOfLastWordFor(String s) {
    if (Objects.isNull(s) || s.length() == 0) {
        return 0;
    }
    int start = s.length(), end = s.length();
    for (int i = s.length() - 1; i >= 0; i--) {
        char c = s.charAt(i);
        if (end == s.length() && c != ' ') {
            end = i;
        }
        if (c == ' ' && i < end && end != s.length()) {
            start = i;
            break;
        }
    }
    if (start == s.length() && end != s.length()) {
        start = -1;
    }
    return end - start;
}