LEETCODE57数组分类,适中级别,题目:969442695
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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;
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