leetcode题解-169. Majority Element && 189. Rotate Array

Majority Element 题目：

Given an array of size n, find the majority element. The majority element is the element that appears more than ⌊ n/2 ⌋ times.You may assume that the array is non-empty and the majority element always exist in the array.

本题就是为了寻找数组中出现次数超过n/2的数字。一种最简单的思路就是将数组排序，然后判断i和i+n/2是否相等即可。代码入下：

    public int majorityElement(int[] nums) {        Arrays.sort(nums);        for(int i=0; i<nums.length; i++){            if(nums[i] == nums[i+nums.length/2])                return nums[i];        }        return 0;    }

这种方法因为使用了排序，而导致效率较低。由于存在某个数出现的次数大于n/2，比其他数字出现次数之和还要打。所以我们可以使用下面这种方法来求解。代码入下：

    public int majorityElement1(int[] num) {        int major=num[0], count = 1;        for(int i=1; i<num.length;i++){            if(count==0){                count++;                major=num[i];            }else if(major==num[i]){                count++;            }else count--;        }        return major;    }

Rotate Array 题目：

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.

本题是实现数组反转的功能。第一种思路就是使用一个大小为tmp的数组来保存后k个数，然后将前面的n-k个数存到后面，最后再把tmp的k个数保存到前面即可。代码入下所示：

    public void rotate(int[] nums, int k) {        if(nums.length <= 1){            return;        }        int step = k % nums.length;        int [] tmp = new int[step];        for(int i=0; i<step; i++)            tmp[i] = nums[nums.length-step+i];        for(int i=nums.length-step-1; i>=0; i--)            nums[i+step] = nums[i];        for(int i=0; i<step; i++)            nums[i] = tmp[i];    }

另外一种思路就是将数组的前面n-k个数反转，再将后面k个数反转，在将数组反转，就可以得到结果。比如[1,2,3,4,5,6,7], k=3. 先变成[4,3,2,1,5,6,7], 再变成[4,3,2,1,7,6,5]，最后再翻转变成[5,6,7,1,2,3,4]. 这种方法的代码如下》

    public void rotate1(int[] nums, int k) {        if(nums.length <= 1){            return;        }        //step each time to move        int step = k % nums.length;        reverse(nums,0,nums.length - 1);        reverse(nums,0,step - 1);        reverse(nums,step,nums.length - 1);    }    //reverse int array from n to m    public void reverse(int[] nums, int n, int m){        while(n < m){            nums[n] ^= nums[m];            nums[m] ^= nums[n];            nums[n] ^= nums[m];            n++;            m--;        }    }