package y2019.Algorithm.array.medium;

import java.util.ArrayList;

import java.util.List;

/**

 * @ProjectName: cutter-point

 * @Package: y2019.Algorithm.array.medium

 * @ClassName: PancakeSort

 * @Author: xiaof

 * @Description: TODO 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.

 *

 * 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.

 *

 * 参考：https://blog.csdn.net/fuxuemingzhu/article/details/85937314

 *

 * @Date: 2019/7/16 8:57

 * @Version: 1.0

 */

public class PancakeSort {

    public List<Integer> solution(int[] A) {

        //思路是这样的，就是每次吧最大的做一个翻转，移动到最前面，然后再翻转到最后面，这样每次都可以从数据中排除掉最大的那个

        //但是由于这个题的数字都是按照1~n的顺序给的值，那么就不需要每次都取最大值，只要取index索引就可以了

        List<Integer> res = new ArrayList<>();

        for(int i = A.length, x; i > 0; --i) {

            //寻找最大的位置

            for(x = 0; A[x] != i; ++x);

            //当x所在的索引跟当前应该的最大值相等的时候，也就是x指向了最大值的位置的-1位置，我们翻转两次

            //第一次吧值翻转到最前面，第二次翻转到最后面

            reverse(A, x);

            res.add(x + 1);

            //然后翻转到对应的位置

            reverse(A, i - 1);

            res.add(i);

        }

        return res;

    }

    private void reverse(int[] A, int k) {

        //翻转k位

        for(int i = 0, j = k; i < j; ++i,--j) {

            //前后交换

            int temp = A[i];

            A[i] = A[j];

            A[j] = temp;

        }

    }

    public static void main(String[] args) {

        int data[] = {3,2,4,1};

        PancakeSort fuc = new PancakeSort();

        System.out.println(fuc.solution(data));

        System.out.println();

    }

}

package y2019.Algorithm.array.medium;

import java.util.ArrayList;

import java.util.Arrays;

import java.util.List;

import java.util.Set;

/**

 * @ProjectName: cutter-point

 * @Package: y2019.Algorithm.array.medium

 * @ClassName: FindDuplicates

 * @Author: xiaof

 * @Description: TODO 442. Find All Duplicates in an Array

 * Given an array of integers, 1 ≤ a[i] ≤ n (n = size of array), some elements appear twice and others appear once.

 * Find all the elements that appear twice in this array.

 * Could you do it without extra space and in O(n) runtime?

 *

 * Input:

 * [4,3,2,7,8,2,3,1]

 * Output:

 * [2,3]

 *

 * 给定一个整数数组 a，其中1 ≤ a[i] ≤ n （n为数组长度）, 其中有些元素出现两次而其他元素出现一次。

 * 找到所有出现两次的元素。

 * 你可以不用到任何额外空间并在O(n)时间复杂度内解决这个问题吗？

 * @Date: 2019/7/16 9:00

 * @Version: 1.0

 */

public class FindDuplicates {

    public List<Integer> solution(int[] nums) {

        //直接用set

        List<Integer> res = new ArrayList<>();

        if(nums == null || nums.length <= 0) {

            return res;

        }

        //受限还是排序

        Arrays.sort(nums);

        //遍历一次

        int preValue = nums[0];

        for(int i = 1; i < nums.length; ++i) {

            int curValue = nums[i];

            if(preValue == curValue) {

                res.add(curValue);

            } else {

                preValue = curValue;

            }

        }

        return res;

    }

}

package y2019.Algorithm.array.medium;

/**

 * @ProjectName: cutter-point

 * @Package: y2019.Algorithm.array.medium

 * @ClassName: MaxAreaOfIsland

 * @Author: xiaof

 * @Description: TODO 695. Max Area of Island

 * Given a non-empty 2D array grid of 0's and 1's, an island is a group of 1's (representing land)

 * connected 4-directionally (horizontal or vertical.) You may assume all four edges of the grid are surrounded by water.

 *Find the maximum area of an island in the given 2D array. (If there is no island, the maximum area is 0.)

 *

 * [[0,0,1,0,0,0,0,1,0,0,0,0,0],

 *  [0,0,0,0,0,0,0,1,1,1,0,0,0],

 *  [0,1,1,0,1,0,0,0,0,0,0,0,0],

 *  [0,1,0,0,1,1,0,0,1,0,1,0,0],

 *  [0,1,0,0,1,1,0,0,1,1,1,0,0],

 *  [0,0,0,0,0,0,0,0,0,0,1,0,0],

 *  [0,0,0,0,0,0,0,1,1,1,0,0,0],

 *  [0,0,0,0,0,0,0,1,1,0,0,0,0]]

 * Given the above grid, return 6. Note the answer is not 11, because the island must be connected 4-directionally.

 *

 * 给定一个包含了一些 0 和 1的非空二维数组 grid , 一个 岛屿 是由四个方向 (水平或垂直) 的 1 (代表土地) 构成的组合。你可以假设二维矩阵的四个边缘都被水包围着。

 * 找到给定的二维数组中最大的岛屿面积。(如果没有岛屿，则返回面积为0。)

 * 来源：力扣（LeetCode）

 * 链接：https://leetcode-cn.com/problems/max-area-of-island

 * 著作权归领扣网络所有。商业转载请联系官方授权，非商业转载请注明出处。

 *

 * @Date: 2019/7/16 9:00

 * @Version: 1.0

 */

public class MaxAreaOfIsland {

    public int solution(int[][] grid) {

        //寻找聚集度最高的和,遍历所有的岛屿，然后对附近的所有1求和,每次求和探索四个位置的和,上下左右

        int maxIsland = 0;

        for(int i = 0; i < grid.length; ++i) {

            for(int j = 0; j < grid[i].length; ++j) {

                //求出最大的岛屿

                maxIsland = Math.max(maxIsland, areaOfIsLand(grid, i, j));

            }

        }

        return maxIsland;

    }

    private int areaOfIsLand(int[][] grid, int x, int y) {

        if(x >= 0 && x < grid.length && y >=0 && y < grid[x].length && grid[x][y] == 1) {

            //设置标记

            grid[x][y] &= 0;

            return 1 + areaOfIsLand(grid, x - 1, y) + areaOfIsLand(grid, x + 1, y) + areaOfIsLand(grid, x, y - 1) + areaOfIsLand(grid, x, y + 1);

        }

        return 0;

    }

}