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【LeetCode】Rotate Array 旋转数组

2015-04-23 15:38 423 查看

题目

Rotate Array

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.

题目大意

旋转数组,给定数组长度n,旋转位移k,将数组元素循环右移k位。

思路

三步翻转法,《编程珠玑》一书上有讲,步骤如下:

将数组从k位置分为前后两半;

翻转前半部分;

翻转后半部分;

翻转整个数组;

图示如下:



解答

Java代码如下,注意遍历时候的边界不要出错:

public void rotate(int[] nums, int k) {
if(k==0 || nums==null ) return;
k = k%nums.length;
int tmp;
for (int i = 0; i < (nums.length-k)/2; i++) {
tmp = nums[i];
nums[i] = nums[nums.length-k-1-i];
nums[nums.length-k-1-i] = tmp;
}
/*for (int i : nums) {
System.out.print(i+"-");
}
System.out.println();*/

for (int i = 0; i < k/2; i++) {
tmp = nums[i+nums.length-k];
nums[i+nums.length-k] = nums[nums.length-1-i];
nums[nums.length-1-i] = tmp;
}

/*for (int i : nums) {
System.out.print(i+" ");
}
System.out.println();*/

for (int i = 0; i < nums.length/2; i++) {
tmp = nums[i];
nums[i] = nums[nums.length-1-i];
nums[nums.length-1-i] = tmp;
}
}
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标签:  LeetCode 算法 面试