Problem : Easy

LeetCode: Given a string s and an integer k, reverse the first k characters for every 2k characters counting from the start of the string. If there are fewer than k characters left, reverse all of them. If there are less than 2k but greater than or equal to k characters, then reverse the first k characters and leave the other as original.

Image from Leetcode

Thought

• Java String is immutable so convert given Sting to a charArray will be a good solution.
• When loop through this charArray, beacuse we need to swap the first k elements of the 2k substring, so define how int i increase in the loop is the key for best performance. It doesn’t always have to be i++;
• Then to define when pointer i reach the closer end of the array by thinking (start+k)-1 and how far end pointer is away from the updated i. (illustrion may talk better)

Solution: Java

//题目的意思其实概括为 每隔2k个反转前k个，尾数不够k个时候全部反转

class Solution {
    public String reverseStr(String s, int k) {
        char[] sArray = s.toCharArray();
        for (int i = 0; i < sArray.length; i += 2 * k) {
            int start = i;
            //这里是判断尾数够不够k个来取决end指针的位置
            int end = Math.min(sArray.length - 1, start + k - 1);
            while (start < end) {
                char temp = sArray[start];
                sArray[start] = sArray[end];
                sArray[end] = temp;
                start++;
                end--;
            }
        }
        return new String(sArray);
    }      
}

For more instructions head over to the LeetCode.