###### tags: `leetcode` `easy` `Two Pointers` `String`
# [557. Reverse Words in a String III](https://leetcode.com/problems/reverse-words-in-a-string-iii/description/)

## Description

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

## Examples
### Example 1:

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

### Example 2:

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

## Constraints:

- $1 \leq s.length \leq 5 \times 10^4$
- `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.

## Code

char* reverseWords(char* s) {
    int size = 0, length = strlen(s);
    char *token, *res;

    // initialize length +1(for the last '\0') as '\0'
    res = (char*)malloc((length + 1) * sizeof(char));
    memset(res, '\0', length + 1);

    // split string by " "(here can not use ' ')
    token = strtok(s, " ");
    while (token != NULL) {
        // from the position we recorded reverses characters within the length
        // of return token
        for (int i = strlen(token) - 1; i >= 0; size++, i--)
            res[size] = token[i];
        token = strtok(NULL, " ");
        // if there still has data in the string, we assign the last character
        // as space and record plus one
        if (token)
            res[size++] = ' ';
    }
    return res;
}

## Complexity

|Space |Time |
|- |- |
|$O(N)$|$O(N)$|

## Result

- Runtime : 12 ms, 55.62%
- Memory : 7.7 MB, 6.34%