有序数组取中值

一、有两个有序数组 查找出中位数，并且输出

　　如：arr1[] = {1, 3, 5, 7, 9, 11, 343, 5645, 56756}; arr2[] = {0, 2, 4, 6, 8, 10}; 输出值为 7.0

 　　1、通过合并取值，时间复杂度o(m+n)

　　　　

private static double findmidnumber(int arr1[], int arr2[]) {
    int len1 = arr1.length;
    int len2 = arr2.length;

    int arr3[] = new int[len1 + len2];

    int i = 0,j = 0, k = 0;
    while (i < len1 && j < len2) {
        if (arr1[i] <= arr2[j]) {
            arr3[k++] = arr1[i];
            i++;
        } else {
            arr3[k++] = arr2[j];
            j++;
        }
    }
    while (i < len1) {
        arr3[k++] = arr1[i++];
    }

    while (j < len2) {
        arr3[k++] = arr2[j++];
    }

    system.out.println("find + " + k);
    if (arr3.length % 2 == 0) {
        return (arr3[(len1 + len2) / 2 - 1] + arr3[(len1 + len2) / 2]) / 2.0;
    } else {
        return arr3[(len1 + len2) / 2] * 1.0;
    }
}

 

　　2、取最大中值,时间o(k)
　　　　

private static double findmidnumber1(int arr1[], int arr2[]) {
        int len1 = arr1.length;
        int len2 = arr2.length;

        int arr3[] = new int[(len1 + len2) / 2 +1];

        int i = 0,j = 0, k = 0, index = 0;
        while (index < arr3.length && j < len2 && i < len1) {
            if (arr1[i] <= arr2[j]) {
                arr3[index++] =arr1[i];
                i ++;
            } else {
                arr3[index++] =arr2[j];
                j++;
            }
            k ++;
        }
        while (index < arr3.length) {
            if ( i < len1) {
                arr3[index++] =arr1[i++];
            }
            if ( j < len2) {
                arr3[index++] =arr2[j++];
            }
            k ++;
        }
        system.out.println("find1 + " + k);
        if (arr3.length % 2 == 1) {
            return (arr3[arr3.length - 2] + arr3[arr3.length - 1] )/ 2.0;
        } else {
            return arr3[arr3.length - 1] * 1.0;
        }

    }

　　3、递归二分查找法，时间复杂度为o(log(m+n)

private static double findmidnumber3(int arr1[], int arr2[]) {
        int len1 = arr1.length;
        int len2 = arr2.length;

        int l = (len1 + len2 + 1) / 2;
        int r = (len1 + len2 + 2) / 2;

        return (findmin(arr1, 0, arr2, 0, l) + findmin(arr1, 0, arr2, 0, r)) / 2.0;
    }

    private static int findmin(int arr1[], int l, int arr2[],int r, int k) {
        if (l > arr1.length - 1) return arr2[r + k -1];
        if (r > arr2.length - 1) return arr1[l + k -1];
        if (k == 1) return math.min(arr1[l], arr2[r]);

        n++;
        int min1 = integer.min_value, min2 = integer.min_value;

        if (l + k/2 -1 < arr1.length) {
            min1 = arr1[l + k/2 -1];
        }
        if (r + k/2 -1 < arr2.length) {
            min2 = arr2[r + k/2 -1];
        }
        return  (min1 < min2) ? findmin(arr1, l + k/2, arr2, r, k - k/2) :
            findmin(arr1, l, arr2, r + k/2, k - k/2);
    }