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  • Reverse Words in a string

    Given an input string, reverse the string word by word.

    For example,
    Given s = "the sky is blue",
    return "blue is sky the".

    Update (2015-02-12):
    For C programmers: Try to solve it in-place in O(1) space.

    click to show clarification.

    Clarification:
    • What constitutes a word?
      A sequence of non-space characters constitutes a word.
    • Could the input string contain leading or trailing spaces?
      Yes. However, your reversed string should not contain leading or trailing spaces.
    • How about multiple spaces between two words?
      Reduce them to a single space in the reversed string.
     
    Analyse: first reverse the entire string, then have wordBegin as the index for the starting position of the word, find the left and right index of a word, reverse from left to right, copy left to right to wordBegin, set ' ' to the next of that word, set correct index for wordBegin.
    Be aware of the empty string, the length of word equals to 1. 
    Runtime: 9ms
     1 class Solution {
     2 public:
     3     void reverseWords(string &s) {
     4         if (s.empty()) return;
     5         reverse(s.begin(), s.end());
     6         
     7         int left = 0, right = 0, wordBegin = 0;
     8         int index = 0;
     9         for (; index < s.size(); index++) {
    10             if (isspace(s[index])) continue;
    11             if (index == 0 || isspace(s[index - 1]))
    12                 left = index;
    13             if (index == s.size() - 1 || isspace(s[index + 1])) right = index;
    14             else continue;
    15             
    16             reverse(s.begin() + left, s.begin() + right + 1);
    17             // leading 0s before the word
    18             if (wordBegin < left) {
    19                 while (left <= right) {
    20                     s[wordBegin++] = s[left++];
    21                 }
    22                 s[wordBegin] = ' ';
    23                 wordBegin++;
    24             } else {
    25                 wordBegin = right + 2;
    26             }
    27         }
    28         wordBegin = wordBegin ? wordBegin : 1;
    29         s.erase(s.begin() + wordBegin - 1, s.end());
    30     }
    31 };
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  • 原文地址:https://www.cnblogs.com/amazingzoe/p/5902327.html
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