题目介绍

题解

题解链接：题解

核心思路：通过二分查找的确定分割点使左右两部分元素数量相等。

class Solution {public double findMedianSortedArrays(int[] nums1, int[] nums2) {int n1 = nums1.length;int n2 = nums2.length;// 确保nums1是较短的数组，以优化二分查找if (n1 > n2) {return findMedianSortedArrays(nums2, nums1);}int k = (n1 + n2 + 1) / 2;int left = 0;int right = n1;while (left < right) {int m1 = (right + left) / 2;int m2 = k - m1;if (nums1[m1] < nums2[m2 - 1]) {left = m1 + 1;} else {right = m1;}}int m1 = left;int m2 = k - m1;// 处理边界情况int c1 = Math.max(m1 <= 0 ? Integer.MIN_VALUE : nums1[m1 - 1],m2 <= 0 ? Integer.MIN_VALUE : nums2[m2 - 1]);if ((n1 + n2) % 2 == 1) {return c1;}int c2 = Math.min(m1 >= n1 ? Integer.MAX_VALUE : nums1[m1],m2 >= n2 ? Integer.MAX_VALUE : nums2[m2]);return (c1 + c2) / 2.0;}
}