【每天一道編程系列-2018.2.18】（Ans）

【題目描述】

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)). 

【題目翻譯】

兩個排序數組，找這兩個排序數組的中位數，時間複雜度爲O(log(m+n)) 

【解題思路】

採用類二分查找算法 

【本題答案】

/**
 * @author yesr
 * @create 2018-02-18 下午11:06
 * @desc
 **/
public class Test0218 {
    /**
     * <pre>
     * 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)).
     *
     * 題目大意：
     * 兩個排序數組，找這兩個排序數組的中位數，時間複雜度爲O(log(m+n))
     *
     * 題解思路：
     * 遞歸分治求解
     * </pre>
     *
     * @param nums1
     * @param nums2
     * @return
     */
    public double findMedianSortedArrays(int[] nums1, int[] nums2) {

        if (nums1 == null) {
            nums1 = new int[0];
        }

        if (nums2 == null) {
            nums2 = new int[0];
        }

        int len1 = nums1.length;
        int len2 = nums2.length;

        if (len1 < len2) {
            // 確保第一個數組比第二個數組長度大
            return findMedianSortedArrays(nums2, nums1);
        }

        // 如果長度小的數組長度爲0，就返回前一個數組的中位數
        if (len2 == 0) {
            return (nums1[(len1 - 1) / 2] + nums1[len1 / 2]) / 2.0;
        }


        int lo = 0;
        int hi = len2 * 2;
        int mid1;
        int mid2;
        double l1;
        double l2;
        double r1;
        double r2;

        while (lo <= hi) {
            mid2 = (lo + hi) / 2;
            mid1 = len1 + len2 - mid2;

            l1 = (mid1 == 0) ? Integer.MIN_VALUE : nums1[(mid1 - 1) / 2];
            l2 = (mid2 == 0) ? Integer.MIN_VALUE : nums2[(mid2 - 1) / 2];

            r1 = (mid1 == len1 * 2) ? Integer.MAX_VALUE : nums1[mid1 / 2];
            r2 = (mid2 == len2 * 2) ? Integer.MAX_VALUE : nums2[mid2 / 2];

            if (l1 > r2) {
                lo = mid2 + 1;
            } else if (l2 > r1) {
                hi = mid2 - 1;
            } else {
                return (Math.max(l1, l2) + Math.min(r1, r2)) / 2;
            }
        }

        return -1;
    }
}