[LeetCode] 186. Reverse Words in a String II

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

Example:

Input:  ["t","h","e"," ","s","k","y"," ","i","s"," ","b","l","u","e"]
Output: ["b","l","u","e"," ","i","s"," ","s","k","y"," ","t","h","e"]

Note: 

• A word is defined as a sequence of non-space characters.
• The input string does not contain leading or trailing spaces.
• The words are always separated by a single space.

Follow up: Could you do it in-place without allocating extra space?

翻转字符串里的单词II。题意是给一个用char array的句子，请翻转单词的顺序。

思路是先翻转整个input数组，然后用快慢指针扫描input，快指针在空格的地方停下，翻转快慢指针之间的字符。

时间O(n)

空间O(n)

Java实现

class Solution {
    public void reverseWords(char[] s) {
        reverse(s, 0, s.length - 1);
        int r = 0;
        while (r < s.length) {
            int l = r;
            while (r < s.length && s[r] != ' ') {
                r++;
            }
            reverse(s, l, r - 1);
            r++;
        }
    }

    private void reverse(char[] s, int i, int j) {
        while (i < j) {
            char temp = s[i];
            s[i] = s[j];
            s[j] = temp;
            i++;
            j--;
        }
    }
}

 

JavaScript实现

/**
 * @param {character[]} s
 * @return {void} Do not return anything, modify s in-place instead.
 */
var reverseWords = function (s) {
    helper(s, 0, s.length - 1);
    let r = 0;
    while (r < s.length) {
        let l = r;
        while (r < s.length && s[r] != ' ') {
            r++;
        }
        helper(s, l, r - 1);
        r++;
    }
};

var helper = function (s, i, j) {
    while (i < j) {
        let temp = s[i];
        s[i] = s[j];
        s[j] = temp;
        i++;
        j--;
    }
}

 

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