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.
这道题因为js有pop和unshift方法而变得格外简单。。。。

var rotate = function(nums, k) {
    //1
    // var num = nums.length;
    // k = k%num;
    // for (var i = 0; i<k; i++) {
    //     nums.unshift(nums.pop());
    // }
    //2
    // var num = nums.length;
    // k = k%num;
    // k = nums.length-k;
    // for (var i = 0; i<k; i++) {
    //     nums.push(nums.shift());
    // }
    //3
    // var num = nums.length;
    // k = k%num;
    // var deleted = nums.splice(num-k,k);
    // for (var i = deleted.length-1; i>=0;i--)
    //     nums.unshift(deleted[i]);
};

但是对于java这样的语言来说就不是了。
我们这样做，[1,2,3,4,5,6,7] k=3
先全反过来
[7,6,5,4,3,2,1]
反前k个
[5,6,7,4,3,2,1]
反后面的
[5,6,7,1,2,3,4]
神奇吧~

var rotate = function(nums, k) {
    var num = nums.length;
    k = k%num;
    nums.reverse();
    for (var i = 0; i < parseInt(k/2); i++) {
        var temp = nums[i];
        nums[i] = nums[k-i-1];
        nums[k-i-1] = temp;
    }
    for (i = k; i < parseInt((k+num)/2); i++) {
        var temp = nums[i];
        nums[i] = nums[k+num-i-1];
        nums[k+num-i-1] = temp;
    }
};