leetcode解题系列:翻转数组

Ｑ：Problem: 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] isrotated to [5,6,7,1,2,3,4]. How many different ways do you know to solve this problem?a：　　1、first method：新建一个数组，将数据保存存放结果。时间复杂度为：ｎ，空间复杂度:n

public static void rotateArray(int array[], int k) {
if(k <= 0)
return;
if(k > array.length) {
k = k % array.length;
}

int result[] = new int[array.length];

for(int i=0; i<k; i++) {
result[i] = array[array.length - k + i];
}

int j =0;
for(int i= k; i<array.length; i++) {
result[i] = array[j];
j++;
}

System.arraycopy(result, 0, array, 0, array.length);
}

2、second method : 类是冒泡算法：时间：n　* k,空间：１

　　　　　　　　

public static void bubbleRotate(int array[], int k) {

for(int i=k; i< array.length; i++) {
int tempValue = array[i];
for(int j = i ; j > 0; j --) {
//                int temp = array[j];
array[j] = array[j -1];
//                array[j-1] = temp;
}
array[0] = tempValue;
}

}

　　３、third method :翻转,时间为　ｎ,空间为　１

Assuming we are given {1,2,3,4,5,6} and order 2. The basic idea is:1. Divide the array two parts: 1,2,3,4 and 5, 62. Rotate first part: 4,3,2,1,5,63. Rotate second part: 4,3,2,1,6,54. Rotate the whole array: 5,6,1,2,3,4

public static void reverseArray(int array[], int k) {reverse(array, 0, k-1);reverse(array, k, array.length -1);reverse(array, 0, array.length -1);}public static void reverse(int array[], int left, int right) {while (left < right) {int temp = array[left];array[left] = array[right];array[right] = temp;right--;left++;}

        ps:贴下代码 https://code.csdn.net/snippets/775491.git