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LeetCode
  • Overview
  • Mixed / Design
    • Two Sum (4 Qs)
      • 0001. Two Sum
      • 0167. Two Sum II - Input array is sorted
      • 0170. Two Sum III - Data structure design
      • 0653. Two Sum IV - Input is a BST
    • 0015. 3Sum
    • 0208. Implement Trie (Prefix Tree)
  • String
    • 0014. Longest Common Prefix
    • 0028. Implement strStr()
    • 0344. Reverse String
    • 0151. Reverse Words in a String
    • 0557. Reverse Words in a String III
    • 0067. Add Binary
    • 0415. Add Strings
    • 0038. Count and Say
    • 0394. Decode String
    • 0953. Verifying an Alien Dictionary
    • 0020. Valid Parentheses
  • Linked List
    • 0002. Add Two Numbers
    • 0445. Add Two Numbers II
    • 0021. Merge Two Sorted Lists
    • 0023. Merge k Sorted Lists
    • 0206. Reverse Linked List
    • 0019. Remove Nth Node From End of List
  • Tree
    • 0098. Validate BST
    • 0100. Same Tree
    • 0101. Symmetric Tree
    • 0226. Invert Binary Tree
    • 0110. Balanced Binary Tree
    • 0250. Count Univalue Subtrees
    • 0654. Maximum Binary Tree
    • Binary Tree Traversal (7 Qs)
      • 0144. Preorder Traversal
      • 0145. Postorder Traversal
      • 0094. Inorder Traversal
      • 0102. Level Order Traversal
      • 0104. Maximum Depth
      • 0111. Minimum Depth
      • 1302. Deepest Leaves Sum
      • 0993. Cousins in Binary Tree
    • N-ary Tree Traversal (4 Qs)
      • 0589. Preorder Traversal
      • 0590. Postorder Traversal
      • 0429. Level Order Traversal
      • 0559. Maximum Depth
    • Convert to BST (2 Qs)
      • 0108. Convert Sorted Array to Binary Search Tree
      • 0109. Convert Sorted List to Binary Search Tree
  • Binary Search
    • 0704. Binary Search
    • 0035. Search Insert Position
    • 0278. First Bad Version
    • 0367. Valid Perfect Square
    • 0069. Sqrt(x)
    • 0875. Koko Eating Bananas
    • 1011. Capacity To Ship Packages Within D Days
    • 0410. Split Array Largest Sum
    • 0004. Median of Two Sorted Arrays
  • Two Pointer
    • 0075. Sort Colors
    • 0088. Merge Sorted Array
    • 0283. Move Zeroes
    • 0125. Valid Palindrome
    • 0011. Container With Most Water
    • 0026. Remove Duplicates from Sorted Array
  • Sliding Window
    • 0003. Longest Substring Without Repeating Characters
  • Sorting
    • 0148. Sort List
    • 0912. Sort an Array
    • 0215. Kth Largest in Array
  • Dynamic Programming
    • 0509. Fibonacci Number
    • 1137. N-th Tribonacci Number
    • 0055. Jump Game
    • 0053. Maximum Subarray
    • 0022. Generate Parentheses
    • 0005. Longest Palindromic Substring
    • 0072. Edit Distance
    • Buy and Sell Stock (6 Qs)
      • 0122. Best Time to Buy and Sell Stock II
      • 0714. Best Time to Buy and Sell Stock with Transaction Fee
      • 0121. Best Time to Buy and Sell Stock
      • 0309. Best Time to Buy and Sell Stock with Cooldown
      • 0123. Best Time to Buy and Sell Stock III
      • 0188. Best Time to Buy and Sell Stock IV
  • DFS / BFS
    • 0200. Number of Islands
  • Math
    • 0007. Reverse Integer
    • 0009. Palindrome Number
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  1. String

0557. Reverse Words in a String III

Easy | String | 28 ms (92.61%), 15.2 MB (10.81%)

Previous0151. Reverse Words in a StringNext0067. Add Binary

Last updated 3 years ago

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Source: GitHub:

Given a string s, reverse the order of characters in each word within a sentence while still preserving whitespace and initial word order.

Constraints:

  • 1 <= s.length <= 5 * 10^4

  • s contains printable ASCII characters.

  • s does not contain any leading or trailing spaces.

  • There is at least one word in s.

  • All the words in s are separated by a single space.

Input: s = "Let's take LeetCode contest"
Output: "s'teL ekat edoCteeL tsetnoc"

Input: s = "God Ding"
Output: "doG gniD"

  1. Extract words from the input string

  2. Reverse the word then append to the returned string

class Solution:
    def reverseWords(self, s: str) -> str:
        
        # ==================================================
        #  String                                          =
        # ==================================================
        # time  : O(n), n is the length of s
        # space : O(n)
        
        l, r = 0, len(s) - 1
        word, ans = '', ''
        while l <= r:
            if s[l] == ' ':
                if word: ans += self.reverse(word) + ' '
                word = ''
                l += 1
                continue
                
            word += s[l]
            l += 1
        
        if word: ans += self.reverse(word) + ' '
            
        return ans[:-1]
            
    def reverse(self, s: str) -> str:
        s, l, r = [char for char in s], 0, len(s) - 1
        while l < r:
            s[l], s[r] = s[r], s[l]
            l += 1
            r -= 1
            
        ret = ''
        for char in s: ret += char
        return ret
    
    '''
    def reverseWords(self, s: str) -> str:
        return ' '.join(x[::-1] for x in s.split())
    '''
class Solution {
    /**
     * @time  : O(n)
     * @space : O(n)
     */
    
    public String reverseWords(String input) {
        StringBuilder result = new StringBuilder();
        StringBuilder word = new StringBuilder();
        
        for (int i = 0; i < input.length(); i++) {
            if (input.charAt(i) != ' ') {
                word.append(input.charAt(i));
            } else {
                result.append(word.reverse());
                result.append(" ");
                word.setLength(0);
            }
        }
        
        result.append(word.reverse());
        return result.toString();
    }
}
LeetCode - Reverse Words in a String III
Solution / Performance