常见排序算法JAVA实现

1、冒泡排序，时间复杂度：最好：T(n) = O(n) ，情况：T(n) = O(n2) ，平均：T(n) = O(n2)

public int[] bubbleSort(int[] nums) {
        if (nums.length < 2) {
            return nums;
        }
        for (int i = 0; i < nums.length; i++) {
            for (int j = 0; j < nums.length - 1 - i; j++) {
                if (nums[j] > nums[j + 1]) {
                    int tem = nums[j];
                    nums[j] = nums[j + 1];
                    nums[j + 1] = tem;
                }
            }
        }
        return nums;
    }

2、选择排序，时间复杂度：最好：T(n) = O(n2) ，最差：T(n) = O(n2) ，平均：T(n) = O(n2)

public int[] selectSort(int[] nums) {
        if (nums.length < 2) {
            return nums;
        }
        for (int i = 0; i < nums.length - 1; i++) {
            int minIndex = i;
            for (int j = i + 1; j < nums.length; j++) {
                if (nums[minIndex] > nums[j]) {
                    minIndex = j;
                }
            }
            if (minIndex != i) {
                int tem = nums[i];
                nums[i] = nums[minIndex];
                nums[minIndex] = tem;
            }
        }
        return nums;
    }

3、插入排序，时间复杂度：最好：T(n) = O(n) ，情况：T(n) = O(n2) ，平均：T(n) = O(n2)

public int[] insertSort(int[] nums) {
        if (nums.length < 2) {
            return nums;
        }
        for (int i = 0; i < nums.length - 1; i++) {
            int cur = nums[i + 1];
            int per = i;
            while (per >= 0 && nums[per] > cur) {
                nums[per + 1] = nums[per];
                per--;
            }
            nums[per + 1] = cur;
        }
        return nums;
    }

 4、快速排序，时间复杂度：最好：T(n) = O(nlogn) ，最差：T(n) = O(n2)， 平均：T(n) = O(nlogn)。它无法保证相等的元素相对位置不变，是不稳定的排序

public int[] quickSort(int[] nums) {
    if (nums.length < 2) {
        return nums;
    }
    sort(nums, 0, nums.length - 1);
    return nums;
}

private void sort(int nums[], int low, int high) {
    int l = low, h = high;
    int povit = nums[low];
    while (l < h) {
        while (l < h && nums[h] >= povit)
            h--;
        if (l < h) {
            nums[l] = nums[h];
            l++;
        }
        while (l < h && nums[l] <= povit)
            l++;
        if (l < h) {
            nums[h] = nums[l];
            h--;
        }
    }
    nums[l] = povit;
    if (l - 1 > low)
        sort(nums, low, l - 1);
    if (h + 1 < high)
        sort(nums, h + 1, high);
}

 5、归并排序，时间复杂度：最好：T(n) = O(n) ，最差：T(n) = O(nlogn) ，平均：T(n) = O(nlogn)

public int[] mergeSort(int[] nums) {
    if (nums.length < 2) {
        return nums;
    }
    int mid = nums.length / 2;
    int[] nums1 = Arrays.copyOfRange(nums, 0, mid);
    int[] nums2 = Arrays.copyOfRange(nums, mid, nums.length);
    return merge(mergeSort(nums1), mergeSort(nums2));
}

private int[] merge(int[] nums1, int[] nums2) {
    int[] back = new int[nums1.length + nums2.length];
    for (int i = 0, m = 0, n = 0; i < back.length; i++) {
        if (m == nums1.length) {
            back[i] = nums2[n++];
        } else if (n == nums2.length) {
            back[i] = nums2[m++];
        } else if (nums1[m] < nums2[n]) {
            back[i] = nums1[m++];
        } else {
            back[i] = nums2[n++];
        }
    }
    return back;
}