每3日一篇leetcode，第四題：MedianOfTwoSortedArrays

1.尋找兩個排序的數組的中位數

//2 ms	47.4 MB,時間複雜度O(n)
    public double findMedianSortedArrays(int[] nums1, int[] nums2) {
        double a = 0, b = 0;
        int size1 = nums1.length;
        int size2 = nums2.length;
        int i = 0, j = 0;
        boolean single = (size1 + size2) % 2 != 0;
        int middle = (size1 + size2) / 2;
        if (size1 == 0) {
            return single ? nums2[middle] : ((double) nums2[middle] + nums2[middle - 1]) / 2;
        } else if (size2 == 0) {
            return single ? nums1[middle] : ((double) nums1[middle] + nums1[middle - 1]) / 2;
        }

        while (i < size1 && j < size2) {
            if (nums1[i] <= nums2[j]) {
                if (single) {
                    if (i + j == middle) {
                        return nums1[i];
                    }
                } else {
                    if (i + j == middle - 1) {
                        a = nums1[i];
                    } else if (i + j == middle) {
                        b = nums1[i];
                        return (a + b) / 2;
                    }
                }
                i++;
            } else {
                if (single) {
                    if (i + j == middle) {
                        return nums2[j];
                    }
                } else {
                    if (i + j == middle - 1) {
                        a = nums2[j];
                    } else if (i + j == middle) {
                        b = nums2[j];
                        return (a + b) / 2;
                    }
                }
                j++;
            }
        }

        if (i == size1) {
            return getMedian(nums1, nums2, size1, single, middle);
        } else if (j == size2) {
            return getMedian(nums2, nums1, size2, single, middle);
        }
        return 0;
    }

    private double getMedian(int[] nums1, int[] nums2, int size1, boolean single, int middle) {
        double b;
        double a;
        if (single) {
            return nums2[middle - size1];
        } else {
            if (middle - size1 > 0) {
                b = nums2[middle - size1];
                a = Math.max(nums2[middle - size1 - 1], nums1[size1 - 1]);
                return (a + b) / 2;
            } else {
                b = nums2[middle - size1];
                a = nums1[size1 - 1];
                return (a + b) / 2;
            }

        }
    }
複製代碼