数组翻转 I am a person => person a am I

package com.study.algorithm;

public class Demo {
    public static void main(String[] args) {
        // 将该数组翻转 输出 person a am I，要求空间复杂度为O(1)，不考虑时间复杂度
        char[] words = {'I', ' ', 'a', 'm', ' ', 'a', ' ', 'p', 'e', 'r', 's', 'o', 'n'};
        print(words);
        // 执行方法
        revert(words);
        print(words);
    }

    private static void revert(char[] arr) {
        // 将数组全部反转
        reverseArr(arr, -1, -1);
        int start = 0, end = 0;
        for (int i = 0; i < arr.length; i++) {
            if (arr[i] == ' ') {
                end = i - 1;
                reverseArr(arr, start, end);
                start = i + 1;
            }
        }
    }

    /**
     * 反转数组
     * @param arr
     * @param index
     * @param to
     */
    private static void reverseArr(char[] arr, int index, int to) {
        System.out.println("index = " + index + " , to = " + to);
        index = index < 0 ? 0 : index;
        to = to < 0 ? arr.length - 1 : to;

        for (int i = 0; i <= (to - index) / 2; i++) {
            char temp = arr[index + i];
            arr[index + i] = arr[to - i];
            arr[to - i] = temp;
        }
    }

    private static void print(char[] arr) {
        for (char c : arr) {
            System.out.print(c);
        }
        System.out.println();
    }
}