排序算法

package practice;/** * @version 2017.10.1 * @author XINGXIN * */public class Sort {    /**     * <p>选择排序O(n^2).<br>     * @param nums: int[].     * @return nums.      */    public int[] selectSort(int[] nums) {        for (int i=0; i<nums.length; i++) {            int minindex = i;            for (int j=i; j<nums.length; j++) {                if (nums[j] < nums[minindex]) {                    minindex = j;                }            }            int temp = nums[i];            nums[i] = nums[minindex];            nums[minindex] = temp;        }        return nums;    }    /**     * <p>冒泡排序O(n^2).<br>     * @param nums: int[].     * @return nums.      */    public int[] bubbleSort(int[] nums) {        for (int i=0; i<nums.length-1; i++) {            for (int j=0; j<nums.length-1-i; j++) {                if (nums[j] > nums[j+1]) {                    int temp = nums[j];                    nums[j] = nums[j+1];                    nums[j+1] = temp;                }            }        }        return nums;    }    /**     * <p>插入排序O(n^2).<br>     * @param nums: int[].     * @return nums.      */    public int[] insertionSort(int[] nums) {        for (int i=1; i<nums.length; i++) {            for (int j=i; j>0; j--) {                if (nums[j] < nums[j-1]) {                    int temp = nums[j];                    nums[j] = nums[j-1];                    nums[j-1] = temp;                }            }        }        return nums;    }    /**     * <p>快速排序O(nlogn).</br>     * 递归.     * @param nums[]: 数组     * @param start: 0     * @param end: nums.length - 1     * @return int[] nums.     */    public int[] quickSort(int[] nums, int start, int end) {        int key = nums[start];  //第一个作为key        int i=start, j=end;        /*key值固定，排一次*/        while (i != j) {            if (nums[j]>=key && i<j)                j--;            nums[i] = nums[j];  //!!不用key进行替换，一次while之后就自动交换比key大和小的值，key的值在最后才替换            if (nums[i]<key && i<j)                i++;            nums[j] = nums[i];        }        nums[i] = key;  //i==j，需要将key替换到i或j的位置        /*排左边*/        if (i-1 > start)            quickSort(nums, start, i-1);        /*排右边*/        if (j+1 < end)            quickSort(nums, j+1, end);        return nums;    }    /**     * <p>归并排序.<br>     * @param nums: int[]     * @param low: 0.     * @param high: nums.length-1.     */    public int[] mergeSort(int[] nums, int low, int high) {        int mid = (low + high) / 2;        if (low < high) {            mergeSort(nums, low, mid);  //左边排序            mergeSort(nums, mid+1, high);   //右边排序            merge(nums, low, mid, high);    //两边连接        }        return nums;    }    /*重点：两个数组如何合并？详细见思想*/    public int[] merge(int[] ns, int low, int mid, int high) {        int[] temp = new int[high-low+1];           int i=low, j=mid+1, k=0;        while (i<=mid && j<=high) {             if (ns[i] <= ns[j]) {   //比较剩下的两个数组的第一个，小的放入temp                temp[k++] = ns[i++];            } else {                temp[k++] = ns[j++];            }        }        //ns前半段还有剩余        while (i <= mid) {            temp[k++] = ns[i++];        }        //ns后半段还有剩余        while (j <= high) {            temp[k++] = ns[j++];        }        for (int k2=0; k2<temp.length; k2++) {            ns[k2+low] = temp[k2];        }        return ns;    }}