969. Pancake Sorting(排序)

package Sort;

import java.util.ArrayList;
import java.util.Arrays;
import java.util.List;

public class PancakeSort_969 {
	// 969. Pancake Sorting
	/*
	 * Given an array A, we can perform a pancake flip: We choose some positive
	 * integer k <= A.length, then reverse the order of the first k elements of A.
	 * We want to perform zero or more pancake flips (doing them one after another
	 * in succession) to sort the array A.
	 * 
	 * Return the k-values corresponding to a sequence of pancake flips that sort A.
	 * Any valid answer that sorts the array within 10 * A.length flips will be
	 * judged as correct.
	 * 
	 * 
	 * 
	 * Example 1:
	 * 
	 * Input: [3,2,4,1] Output: [4,2,4,3] Explanation: We perform 4 pancake flips,
	 * with k values 4, 2, 4, and 3. Starting state: A = [3, 2, 4, 1] After 1st flip
	 * (k=4): A = [1, 4, 2, 3] After 2nd flip (k=2): A = [4, 1, 2, 3] After 3rd flip
	 * (k=4): A = [3, 2, 1, 4] After 4th flip (k=3): A = [1, 2, 3, 4], which is
	 * sorted.
	 * 
	 * Example 2:
	 * 
	 * Input: [1,2,3] Output: [] Explanation: The input is already sorted, so there
	 * is no need to flip anything. Note that other answers, such as [3, 3], would
	 * also be accepted.
	 * 
	 * 
	 * 
	 * Note:
	 * 
	 * 1 <= A.length <= 100 A[i] is a permutation of [1, 2, ..., A.length]
	 * 
	 * 來源：力扣（LeetCode） 鏈接：https://leetcode-cn.com/problems/pancake-sorting
	 * 著作權歸領釦網絡所有。商業轉載請聯繫官方授權，非商業轉載請註明出處。
	 */

	// 思路：
	// 如果想把某個數移到你想到的位置，分兩步：
	// 如：[3,2,4,1] 把4移動到最後面
	// k=3 --》把4移動到首位
	// k=4 --》把4移動到最後
	// 因此從i=A.length()-1開始，依次挑選最大的元素放在i位置
	public static void main(String[] args) {
		PancakeSort_969 p = new PancakeSort_969();
		int[] A = new int[] { 3, 2, 4, 1 };
		List<Integer> list = p.pancakeSort(A);
	}

	public List<Integer> pancakeSort(int[] A) {
		List<Integer> list = new ArrayList<Integer>();
		for (int i = A.length - 1; i > 0; i--) {
			int index = max_index(A, i);
			if (index > 0) {
				reverse(A, index);
				list.add(index + 1);
			}
			reverse(A, i);
			list.add(i + 1);
		}

		return list;

	}

	private void reverse(int[] a, int index) {
		for (int i = 0; i <= index / 2; i++) {
			int tmp = a[i];
			a[i] = a[index - i];
			a[index - i] = tmp;
		}

	}

	private int max_index(int[] a, int i) {
		int index = 0;
		for (int k = 1; k <= i; k++) {
			if (a[k] > a[index]) {
				index = k;
			}
		}
		return index;
	}

}