冒泡排序及其优化代码

初级：

	public static int[] sortMaoPao1(int[] arr){
        int count1 = 0, count2 = 0;
        int len = arr.length;
        for (int i = 0; i < len -1; i++) {
            for (int j = 0; j < len - 1 - i; j++) {
                // 从小到大排序
                if(arr[j + 1] < arr[j]){
                    System.out.println("交换了：" + arr[j] + "和" + arr[j+1]);
                    int tmp = arr[j];
                    arr[j] = arr[j+1];
                    arr[j+1] = tmp;
                    count1++;
                }
                count2 ++;
               /* // 从大到小排序
                if(arr[j + 1] > arr[j]){
                    int tmp = arr[j];
                    arr[j] = arr[j+1];
                    arr[j+1] = tmp;
                }*/
            }
        }
        System.out.println("count1:" + count1);     // 5
        System.out.println("count2:" + count2);     // 45 = 9+8+7+6+5+4+3+2+1,其中arr.length = 10
        return arr;
    }

中级：

	public static int[] sortMaoPao2(int[] arr){
        int count1 = 0, count2 = 0;
        int len = arr.length;
        for (int i = 0; i < len -1; i++) {
            boolean flag = true;
            for (int j = 0; j < len - 1 - i; j++) {
                // 从小到大排序
                if(arr[j + 1] < arr[j]){
                    System.out.println("交换了：" + arr[j] + "和" + arr[j+1]);
                    int tmp = arr[j];
                    arr[j] = arr[j+1];
                    arr[j+1] = tmp;
                    count1++;
                    flag = false;
                }
                count2++;
               /* // 从大到小排序
                if(arr[j + 1] > arr[j]){
                    int tmp = arr[j];
                    arr[j] = arr[j+1];
                    arr[j+1] = tmp;
                }*/

            }
            // flag == true,说明 if(arr[j + 1] < arr[j]) 程序未执行，即不会改变flag，所以直接break
            if(flag == true){
                break;
            }
        }
        System.out.println("count1:" + count1);     // 5
        System.out.println("count2:" + count2);     // 30 = 9+8+7+6,循环次数明显减少
        return arr;
    }

高级：

	public static int[] sortMaoPao3(int[] arr){
        int count1 = 0, count2 = 0;
        int lastOutOfIndex = 0;
        int len = arr.length;
        int sortBorder = len - 1;
        for (int i = 0; i < len -1; i++) {
            boolean flag = true;
            for (int j = 0; j < sortBorder; j++) {
                // 从小到大排序
                if(arr[j + 1] < arr[j]){
                    System.out.println("交换了：" + arr[j] + "和" + arr[j+1]);
                    int tmp = arr[j];
                    arr[j] = arr[j+1];
                    arr[j+1] = tmp;
                    count1++;
                    flag = false;
                    lastOutOfIndex = j;
                }
                count2++;
               /* // 从大到小排序
                if(arr[j + 1] > arr[j]){
                    int tmp = arr[j];
                    arr[j] = arr[j+1];
                    arr[j+1] = tmp;
                }*/

            }
            sortBorder = lastOutOfIndex;
            // flag == true,说明 if(arr[j + 1] < arr[j]) 程序未执行，即不会改变flag，所以直接break
            if(flag == true){
                break;
            }
        }
        System.out.println("count1:" + count1);     // 5
        System.out.println("count2:" + count2);     // 20,循环次数明显减少
        return arr;
    }

测试实例：

	public static void main(String[] args) {
        int[] arr = {1,4,2,6,7,3,23,12,63,171};
        arr = sortMaoPao3(arr);
        System.out.println(Arrays.toString(arr));
    }

理论请参考：图解冒泡