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  • LeetCode Rotate Array

    Rotate Array Total Accepted: 12759 Total Submissions: 73112 My Submissions Question Solution
    Rotate an array of n elements to the right by k steps.

    For example, with n = 7 and k = 3, the array [1,2,3,4,5,6,7] is rotated to [5,6,7,1,2,3,4].

    Note:
    Try to come up as many solutions as you can, there are at least 3 different ways to solve this problem.

    题意:循环数组,n代表数组的长度,k代表向右移动的次数。
    解法一:

    class Solution {
    public:
        void rotate(int nums[], int n, int k) {
            if(n==0)return;
            k=k%n;//当k大于n的时候。n次循环会回到初始位置,因此,能够省略若干次
            if (k == 0) return;  
            int *s=new int[k];//为了一步到位的展开移动,申请k个额外空间用于保存被移出去的元素
            for(int i=0;i<k;++i)
                s[i]=nums[n-k+i];//保存被移出去的元素
            for(int j=n-k-1;j>=0;--j)
                nums[j+k]=nums[j];//移动
            for(int i=0;i<k;++i)
                nums[i]=s[i];//被移出的元素进行归位
            free(s);
        }
    };

    须要额外空间O(k%n)
    33 / 33 test cases passed.
    Status: Accepted
    Runtime: 29 ms

    解法二(网络获取):
    三次翻转法,第一次翻转前n-k个。第二次翻转后k个,第三次翻转所有。

    class Solution {
    public:
        void rotate(int nums[], int n, int k) {
            if(n==0)return ;
            k=k%n;
            if(k==0)return ;
            reverse(nums,n-k,n-1);
            reverse(nums,0,n-k-1);
            reverse(nums,0,n-1);
        }
        void reverse(int nums[],int i,int j)
        {
            for(;i<j;++i,--j)
            {
                int t=nums[i];
                nums[i]=nums[j];
                nums[j]=t;
            }
        }
    };

    33 / 33 test cases passed.
    Status: Accepted
    Runtime: 26 ms

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  • 原文地址:https://www.cnblogs.com/yxwkf/p/5284978.html
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