LeetCode 4. Median of Two Sorted Arrays

원본 사이트 주소:https://leetcode.com/problems/median-of-two-sorted-arrays/
There are two sorted arrays nums1 and nums2 of size m and n respectively. Find the median of the two sorted arrays. The overall run time complexity should be O(log (m+n)).
방법: 이분법으로 찾기.

public class Solution {
    private int median(int[] nums1, int[] nums2, int k) {
        if (nums1.length == 0) return -1;
        if (nums2.length == 0) return k;
        int i=0, j=nums1.length-1;
        while (i<=j) {
            int m = (i+j)/2;
            int n = k-m;
            if (n < 0) j=m-1;
            else if (n>nums2.length) i=m+1;
            else if (n == nums2.length) {
                if (nums2[n-1] <= nums1[m]) return m;
                i=m+1;
            } else if (n == 0) {
                if (nums1[m] <= nums2[n]) return m;
                j=m-1;
            } else {
                if (nums2[n-1] <= nums1[m] && nums1[m] <= nums2[n]) return m;
                if (nums2[n-1] > nums1[m]) i=m+1;
                if (nums1[m] > nums2[n]) j=m-1;
            }
        }
        return -1;
    }
    public double findMedianSortedArrays(int[] nums1, int[] nums2) {
        int n = nums1.length+nums2.length;
        if (n % 2 == 0) {
            int m1 = median(nums1, nums2, n/2-1);
            int median = 0;
            if (m1 != -1) median = nums1[m1]; else median = nums2[median(nums2, nums1, (n-1)/2)];
            int m2 = median(nums1, nums2, n/2);
            if (m2 != -1) median += nums1[m2]; else median += nums2[median(nums2, nums1, n/2)];
            return (double)median/2;
        } else {
            int m = median(nums1, nums2, (n-1)/2);
            if (m != -1) return nums1[m];
            return nums2[median(nums2, nums1, (n-1)/2)];
        }
    }
}

방법2: 귀속.

public class Solution {
    private int find(int[] a, int af, int at, int[] b, int k) {
        if (af > at) return -1;
        if (b.length == 0) {
            if (af <= k && k <= at) return k;
            return -1;
        }
        int m = (af + at) / 2;
        int n = k - m;
        if (n == b.length) {
            if (b[n-1] <= a[m]) return m;
            return find(a, m + 1, at, b, k);
        }
        if (n == 0) {
            if (a[m] <= b[n]) return m;
            return find(a, af, m - 1, b, k);
        }
        if (b[n - 1] <= a[m] && a[m] <= b[n]) return m;
        if (b[n - 1] > a[m]) return find(a, m + 1, at, b, k);
        return find(a, af, m - 1, b, k);
    }
    
    private int kth(int[] nums1, int[] nums2, int k) {
        for(int i = 0; i < 2; i ++) {
            int af = Math.max(0, k - nums2.length);
            int at = Math.min(k, nums1.length - 1);
            int m = find(nums1, af, at, nums2, k);
            if (m != -1) return nums1[m];
            int[] temp = nums1;
            nums1 = nums2;
            nums2 = temp;
        }
        return 0;
    }
    
    public double findMedianSortedArrays(int[] nums1, int[] nums2) {
        if ((nums1.length + nums2.length) % 2 == 1) {
            int k = (nums1.length + nums2.length) / 2;
            return kth(nums1, nums2, k);
        } else {
            int k = (nums1.length + nums2.length) / 2 - 1;
            return (double)(kth(nums1, nums2, k) + kth(nums1, nums2, k + 1)) / 2;
        }
    }
}