标签：turn find cut solution while EDA public and 答案
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)).
You may assume nums1 and nums2 cannot be both empty.
Example 1:
nums1 = [1, 3] nums2 = [2] The median is 2.0
Example 2:
nums1 = [1, 2] nums2 = [3, 4] The median is (2 + 3)/2 = 2.5
第一次没有看答案写出了faster than 99.3%的answer。。不过debug了很多很多次
但是我的time是O(m+n），不符合题意。
class Solution { public double findMedianSortedArrays(int[] nums1, int[] nums2) { int l = nums1.length + nums2.length; int[] mer = new int[l]; int i = 0; int j = 0; int i1 = 0; int j1 = 0; if(nums1.length == 0) { mer = nums2; } else if(nums2.length == 0) { mer = nums1; }else { for(int k = 0; k<l; k++) { if(i1==0 && j1==0){ if(nums1[i]<=nums2[j]) { mer[k] = nums1[i]; if(i==nums1.length-1){i1 = 1;} i++; }else{ mer[k] = nums2[j]; if(j==nums2.length-1){ j1 = 1;} j++; } }else if(i1 == 1){ mer[k] = nums2[j]; if(j<nums2.length-1) j++; }else if(j1 == 1){ mer[k] = nums1[i]; if(i<nums1.length-1) i++; } } } double r; if(l%2 == 0) { r = ((double)mer[l/2-1]+(double)mer[l/2])/2.0; } else { r = mer[(l-1)/2]; } return r; } }
正确答案：
cut + binary search
public double findMedianSortedArrays(int[] nums1, int[] nums2) { int n1 = nums1.length; int n2 = nums2.length; if(n1<n2) return findMedianSortedArrays(nums2,nums1); int lo = 0, hi = 2*n2; while(lo<=hi) { int md2 = (lo+hi) / 2; int md1 = n1 + n2 - md2; double l1 = (md1 == 0)? - Math.pow(2,31) : nums1[(md1-1)/2]; //java里面乘方不能用^ ！！ double l2 = (md2 == 0)? - Math.pow(2,31) : nums2[(md2-1)/2]; double r1 = (md1 == 2*n1)? Math.pow(2,31)-1 : nums1[md1/2]; double r2 = (md2 == 2*n2)? Math.pow(2,31)-1 : nums2[md2/2]; if(l1>r2) { lo = md2+1; }else if(l2>r1) { hi = md2-1; }else { return (Math.max(l1,l2)+Math.min(r1,r2))/2; } } return -1; }
leetcode 4. Median of Two Sorted Arrays
标签：turn find cut solution while EDA public and 答案
原文地址：https://www.cnblogs.com/jamieliu/p/10301607.html
