标签：
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.
Related problem: Reverse Words in a String II
分析：题意为 将n个元素的数组向右旋转k步
思路：用vector容器的东西来做很简单
代码如下：（O(1) Space）
class Solution {
public:
void rotate(vector<int>& nums, int k)
{
for(int i=0;i<k;i++)
{
nums.insert(nums.begin(), nums.back());
nums.pop_back();
}
}
};
然后换种方法：
class Solution {
public:
void rotate(vector<int>& nums, int k) {
int n=nums.size();
vector<int> v(n);
for(int i=n-k,j=0;i<n-1,j<k-1;i++,j++){
v[j]=nums[i];
}
for(int i=0,j=k;i<n-k-1,j<n-1;i++,j++){
v[j]=nums[i];
}
nums=v;
}
};
出错：Last executed input:[1,2,3,4,5,6], 11
因为没有考虑到当k大于n的情况，所以需要改进：
class Solution {
public:
void rotate(vector<int>& nums, int k) {
int n = nums.size();
vector<int> rot(n);
for(int i = 0; i < n; i++) {
if((i + k) < n) rot[i + k] = nums[i];
if((i + k) >= n) {
rot[(i + k)%n] = nums[i];
}
}
nums = rot;
}
};
c语言
看看：
void rotate(int* nums, int numsSize, int k) {
int i;
if(k > numsSize)
k -= numsSize;
int* temp = (int*)calloc(sizeof(int), numsSize);
for(i = 0; i < k; i++)
temp[i] = nums[numsSize - k + i];
for(; i < numsSize; i++)
temp[i] = nums[i - k];
for(i = 0; i < numsSize; i++)
nums[i] = temp[i];
}
或：
void reverse(int *nums, int start, int end) {
int tmp;
while (start < end) {
tmp=nums[start];
nums[start]=nums[end];
nums[end]=tmp;
++start;
--end;
}
}
void rotate(int* nums, int numsSize, int k) {
k=k%numsSize;
if (k==0) return;
reverse(nums,0,numsSize-k-1);
reverse(nums,numsSize-k,numsSize-1);
reverse(nums,0,numsSize-1);
}
标签：
原文地址：http://www.cnblogs.com/carsonzhu/p/4658520.html
