【算法刷題】2：尋找兩個有序數組的中位數

二分搜索的擴展

題目描述

給定兩個大小爲m和n的有序數組num1和num2.請你找出這兩個數組的中位數，而且要求算法的時間複雜度爲O(log(m+n)).你能夠假設num1和num2不會同時爲空。示例：

nums1 = [1, 3]
nums2 = [2]
則中位數是 2.0

nums1 = [1, 2]
nums2 = [3, 4]
則中位數是 (2 + 3)/2 = 2.5
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思路與實現

合併後取中位數

先像歸併排序同樣將兩數組 merge 爲一個有序數組，直接取中位數便可：

private double findMedianSortedArrays(int[] A, int[] B) {
        int m = A.length;
        int n = B.length;
        if (m > n) { // 確保m<=n
            int[] temp = A;
            A = B;
            B = temp;
            int tmp = m;
            m = n;
            n = tmp;
        }
        int iMin = 0, iMax = m, halfLen = (m + n + 1) / 2;
        while (iMin <= iMax) {
            int i = (iMin + iMax) / 2;
            int j = halfLen - i;
            if (i < iMax && B[j - 1] > A[i]) {
                iMin = i + 1; // i is too small
            } else if (i > iMin && A[i - 1] > B[j]) {
                iMax = i - 1; // i is too big
            } else { // i is perfect
                int maxLeft = 0;
                if (i == 0) {
                    maxLeft = B[j - 1];
                } else if (j == 0) {
                    maxLeft = A[i - 1];
                } else {
                    maxLeft = Math.max(A[i - 1], B[j - 1]);
                }

                int minRight = 0;
                if (i == m) {
                    minRight = B[j];
                } else if (j == n) {
                    minRight = A[i];
                } else {
                    minRight = Math.min(B[j], A[i]);
                }

                return (maxLeft + minRight) / 2.0;
            }
        }
        return 0.0;
    }
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