旋转数组
给定一个整数数组 nums,将数组中的元素向右轮转 k 个位置,其中 k 是非负数。
示例 1:
输入: nums = [1,2,3,4,5,6,7], k = 3
输出: [5,6,7,1,2,3,4]
解释:
向右轮转 1 步: [7,1,2,3,4,5,6]
向右轮转 2 步: [6,7,1,2,3,4,5]
向右轮转 3 步: [5,6,7,1,2,3,4]
示例 2:
输入:nums = [-1,-100,3,99], k = 2
输出:[3,99,-1,-100]
解释:
向右轮转 1 步: [99,-1,-100,3]
向右轮转 2 步: [3,99,-1,-100]
我的代码 (看的题解)
public static void round() {
int a[] = new int[]{1,2,3,4,5,6,7};
int k = 3;
reverse(a, 0, a.length - 1);
reverse(a,0,k-1);
reverse(a, k, a.length - 1);
for (int i : a) { System.out.println(i); }
}
public static void reverse(int[] a, int start, int end) {
while (start < end) {
int mid = a[start];
a[start] = a[end];
a[end] = mid;
start++;
end--;
}
}
这个是我看数据结构想的挺有意思的;从后向前遍历,前面的元素向后移一位
int a[] = new int[]{1,2,3,4,5,6,7};
int b = a[a.length - 1];
for (int j = 0; j < 3; j++) {
for (int i = a.length - 2; i >= 0; i--) {
a[i + 1] = a[i];
}
a[0] = b;
b = a[a.length - 1];
}