题目地址:https://leetcode-cn.com/problems/median-of-two-sorted-arrays/
难度:困难
基本思路:
- 将两个有序链表合并,然后根据中位数定义查找
- 分段折半查找
合并链表(low的一笔)
public class Solution { public static void main(String[] args) { Solution s = new Solution(); double ans = s.findMedianSortedArrays(new int[]{1,3,4}, new int[]{2}); System.out.println(ans); } public double findMedianSortedArrays(int[] nums1, int[] nums2) { // 双指针合并数组 int len1 = nums1.length, len2 = nums2.length, n = len1 + len2; int[] res = new int[n]; int p = 0, q = 0, k = 0; if(len1 == 0 && len2 != 0){ res = nums2; }else if(len1 != 0 && len2 == 0){ res = nums1; }else { while (p < len1 && q < len2) { if (nums1[p] > nums2[q]) { res[k++] = nums2[q++]; } else if (nums1[p] == nums2[q]) { res[k++] = nums2[q++]; res[k++] = nums1[p++]; } else { res[k++] = nums1[p++]; } } while (p < len1) { res[k++] = nums1[p++]; } while (q < len2) { res[k++] = nums2[q++]; } int i = 0; while (i++ < res.length) { System.out.print(res[i - 1] + ","); } } if (res.length % 2 == 0) { return (double) (res[res.length / 2] + res[(res.length / 2) - 1]) / 2; } else { return (double) res[(res.length - 1) / 2]; } } }
分段折半查找(今天没想明白,想出来更新)