【Leetcode】4. 寻找两个有序数组的中位数（Median of Two Sorted Arrays）

Leetcode - 4 Median of Two Sorted Arrays (Hard)

题目描述：要求时间复杂度为 O(log(m + n))。

nums1 = [1, 3]
nums2 = [2]

The median is 2.0

解题思路：二分。

      left_part          |        right_part
A[0], A[1], ..., A[i-1]  |  A[i], A[i+1], ..., A[m-1]
B[0], B[1], ..., B[j-1]  |  B[j], B[j+1], ..., B[n-1]

public double findMedianSortedArrays(int[] A, int[] B) {
    int m = A.length, n = B.length;
    if (m > n) { // 确保 n >= m
        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;
        } else if (i > iMin && A[i - 1]  > B[j]) {
            iMax = i - 1;
        } else {
            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]);
            if ((m + n) % 2 == 1) return maxLeft;
            
            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;
}

标签：...,Arrays,minRight,iMin,Median,Two,maxLeft,else,int
来源： https://blog.csdn.net/qq_27124771/article/details/90738751