Partition Array by Odd and Even | 刷题

Partition Array by Odd and Even | 刷题
Partition an integers array into odd number first and even number second. Example Given [1, 2, 3, 4], return [1, 3, 2, 4] Challenge Do it in-place.

    public void partitionArray(int[] nums) {
        int start=0;
        int cur=0;
        int end=nums.length-1;

        while(cur<end){
            if(nums[cur]%2!=0){
                swap(nums, start++, cur++);
            }
            else{
                swap(nums, cur, end--);
            }
        }
    }

http://cherylintcode.blogspot.com/2015/06/partition-array-by-odd-and-even.html
    public void partitionArray(int[] nums) {
        // write your code here;
        if(nums == null || nums.length == 0) return;
       
        int left = 0, right = nums.length - 1;
        while(left < right){
            while(left < right && nums[left] % 2 == 1) left++;
            while(left < right && nums[right] % 2 == 0) right--;
            if(left < right){
                int temp = nums[left];
                nums[left] = nums[right];
                nums[right] = temp;
                left++;
                right--;
            }
           
        }
    }
https://codesolutiony.wordpress.com/2015/05/25/lintcode-partition-array-by-odd-and-even/

    public void partitionArray(int[] nums) {

        if (nums == null || nums.length <= 1) {

            return;

        }

        int even = nums.length;

        for (int i = 0; i < even; i++) {

            if (nums[i] % 2 == 0) {

                int tmp = nums[i];

                nums[i] = nums[--even];

                nums[even] = tmp;

                i--;

            }

        }

    }

http://www.chenguanghe.com/partition-array-by-odd-and-even/

其实就是一个quicksort的partition的改写, 改写的地方很少, 就是判断条件从 nums[i] 对 nums[k]的比较改成nums[i] % 2 != 0的比较.

public void partitionArray(int[] nums) {

        if(nums.length == 0 || nums == null)

            return;

        int i = 0;

        int j = nums.length - 1;

        while(i < j) {

            while(i<j && nums[i] % 2 != 0) // 如果左指针的位置是奇数

                i++;

            while(i< j && nums[j] % 2 == 0) // 如果右指针的位置是偶数

                j--;

            if(i < j)

                swap(nums,i++,j--);

        }

    }

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