标签：over 时间复杂度 题目 2.0 lse bsp find 归并排序 ted
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)). 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
题目要求time complexity 是O(log (m+n))， 先归并排序也是不可行的, 时间复杂度为O(m+n)
class Solution {
public double findMedianSortedArrays(int[] nums1, int[] nums2) {
int l1 = nums1.length;
int l2 = nums2.length;
int i1 = 0;
int i2 = 0;
int i = 0;
int[] nums = new int[l1+l2];
while(i1 < l1 && i2 <l2){
int num1 = nums1[i1];
int num2 = nums2[i2];
if(num1 < num2){
nums[i] = num1;
i1++;
}
else{
nums[i] = num2;
i2++;
}
i++;
}
if(i1 == l1){
while(i2 < l2){
nums[i]=nums2[i2];
i2++;
i++;
}
}
else if(i2 == l2){
while(i1 <l1){
nums[i] = nums1[i1];
i1++;
i++;
}
}
//find median
if(i % 2 != 0){
return (double)nums[i/2];
}
else{
return ((double)nums[i/2-1] + (double)nums[i/2])/2;
}
}
}
LeetCode - Median of Two Sorted Arrays
标签：over 时间复杂度 题目 2.0 lse bsp find 归并排序 ted
原文地址：https://www.cnblogs.com/incrediblechangshuo/p/9065502.html
