}String[] arr = s.split(" ");returnarr[arr.length-1].length(); } 04 第三种解法 将原字符串首尾的空格去掉,然后找到最后一次出现空格的位置,两者相减再减1即为最后单词的长度。 public intlengthOfLastWord3(String s) { return s.trim().length()-s.trim().lastIndexOf(" ")-1; } 05 小结...
leetcode 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...
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. For example, Given s...
Can you solve this real interview question? Length of Last Word - Given a string s consisting of words and spaces, return the length of the last word in the string. A word is a maximal substring consisting of non-space characters only. Example 1:
Leetcode No.58 Length of Last Word最后一个单词的长度(c++实现),1.题目1.1英文题目Givenastringsconsistsofsomewordsseparatedbyspaces,returnthelengthofthelastwordinthestring.Ifthelastworddoesno...
08移除元素Remove Element【Python刷LeetCode 力扣】 09:50 09实现 strStr()Implement strStr()【Python刷LeetCode 力扣】 05:36 10搜索插入位置 Search Insert Position【Python刷LeetCode 力扣】 05:32 11最后一个单词的长度 Length of Last Word【Python刷LeetCode 力扣】 03:42 12加一 Plus One【...
public static int lengthOfLastWord(String s) { s = s.trim(); // 去掉字符串首尾的空格 int i = 0; for (i = s.length() - 1; i >= 0; --i) { if (s.substring(i, i + 1).equals(" ")) { return s.length() - i - 1; ...
LeetCode 58: Length of Last Word 題目描述 題目說明:給一個字串,裡面可能包含英文字母(alphabets) 與空格(space)。回傳最後一個單字(word)的長度。 例如: "Hello World ", 應回傳 5, 因為 World 的長度為 5。 TDD 開始 Step 1, 新增測試用例。長度為2, 不含空格的 word。 s 為"bc" ...
leetcode 58. Length of Last Word 题目描述: 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. No......
public int lengthOfLastWord(String s) { return s.trim().split(" +")[s.trim().split(" +").length - 1].length(); } } 双指针法 复杂度 时间O(N) 空间 O(1) 思路 从后往前看字符串,跳过所有空格后,记下该结束位置,再到下一个空格,再记录一个开始位置,则长度就是结束位置减去开始位置。