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  • LeetCode 969. Pancake Sorting

    原题链接在这里:https://leetcode.com/problems/pancake-sorting/

    题目:

    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. 1 <= A.length <= 100
    2. A[i] is a permutation of [1, 2, ..., A.length]

    题解:

    Find the max index up to largest.

    swap [0, max index]

    swap [0, largest]

    repeat n times.

    Time Complexity: O(n^2). n = A.length.

    Space: O(1).

    AC Java:

     1 class Solution {
     2     public List<Integer> pancakeSort(int[] A) {
     3         List<Integer> res = new ArrayList<>();
     4         if(A == null){
     5             return res;
     6         }
     7         
     8         int n = A.length;
     9         int largest = n;
    10         for(int i = 0; i<n; i++){
    11             int maxIndex = -1;
    12             int max = Integer.MIN_VALUE;
    13             
    14             for(int j = 0; j<largest; j++){
    15                 if(A[j] >= max){
    16                     maxIndex = j;
    17                     max = A[j];
    18                 }    
    19             }
    20             
    21             swap(A, maxIndex);
    22             swap(A, largest-1);
    23             res.add(maxIndex + 1);
    24             res.add(largest--);
    25         }
    26         
    27         return res;
    28     }
    29     
    30     private void swap(int [] arr, int k){
    31         for(int i = 0, j = k; i<j; i++, j--){
    32             int temp = arr[i];
    33             arr[i] = arr[j];
    34             arr[j] = temp;
    35         }
    36     }
    37 }
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  • 原文地址:https://www.cnblogs.com/Dylan-Java-NYC/p/12047200.html
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