004. Median of Two Sorted Arrays
难度Hard
刷题内容
原题连接
- https://leetcode.com/problems/median-of-two-sorted-arrays/submissions/
内容描述
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
思路1 **- 时间复杂度: O(n + m)*- 空间复杂度: O(1)***
直接用暴利搜索，类似与归并两个有序的数组。遍历两个数组，当总长度等于（m+n）/ 2，注意区分总长度奇数和偶数
class Solution {
public:
double findMedianSortedArrays(vector<int>& nums1, vector<int>& nums2) {
int temp = (nums1.size() + nums2.size()) / 2,count1 = 0,i = 0,j = 0,current,pre;
while(i < nums1.size() && j < nums2.size() && count1 <= temp)
{
pre = current;
if(nums1[i] > nums2[j])
current = nums2[j++];
else
current = nums1[i++];
++count1;
}
if(count1 <= temp)
{
if(i < nums1.size())
while(count1 <= temp)
{
pre = current;
current = nums1[i++];
++count1;
}
else
while(count1 <= temp)
{
pre = current;
current = nums2[j++];
++count1;
}
}
if((nums1.size() + nums2.size()) % 2)
return current;
double ans = (current + pre) / 2.0;
return ans;
}
};
思路2 **- 时间复杂度: O(lg(min(n.m)))*- 空间复杂度: O(1)***
我们可以通过二分查找优化算法，利用中位数的定义，将两个数组划分为左右两个部分，nums1左半部分加nums2左半部分等于nums1右半部分加nums2的右半部分，如果总长度为偶数，那么nums1左半部分加nums2左半部分等于nums1右半部分加nums2的右半部分加1。并且max(nums1[i],nums2[j]) <= max(nums1[i + 1],nums2[j + 1])，接下来我们只要二分查找找i，并且要注意边界情况
class Solution {
public:
double findMedianSortedArrays(vector<int>& nums1, vector<int>& nums2) {
int m = nums1.size(),n = nums2.size(),sum = m + n;
if(!nums1.size())
return sum % 2 ? nums2[sum / 2] : (nums2[sum /2] + nums2[sum / 2 - 1]) / 2.0;
if(!nums2.size())
return sum % 2 ? nums1[sum / 2] : (nums1[sum /2] + nums1[sum / 2 - 1]) / 2.0;
if(m > n)
return findMedianSortedArrays(nums2,nums1);
int l = 0,r = m - 1;
while(l < r)
{
int mid = (l + r) / 2;
int j = (sum + 1) / 2 - mid - 2;
int min1 = max(nums1[mid],nums2[j]),max1 = min(nums1[mid + 1],nums2[j + 1]);
if(min1 <= max1)
return sum % 2 ? min1 : (min1 + max1) / 2.0;
else if(nums1[mid] > nums2[j])
r = mid - 1;
else
l = mid + 1;
}
int j = (sum + 1) / 2 - l - 2;
int min1,max1;
if(j < 0)
min1 = nums1[l];
else
min1 = max(nums1[l],nums2[j]);
if(l == nums1.size() - 1)
max1 = nums2[j + 1];
else
max1 = min(nums1[l + 1],nums2[j + 1]);
if(min1 <= max1)
return sum % 2 ? min1 : (min1 + max1) / 2.0;
j++;
if(j < nums2.size() - 1)
max1 = min(nums1[l],nums2[j + 1]);
else
max1 = nums1[l];
min1 = nums2[j];
return sum % 2 ? min1 : (min1 + max1) / 2.0;
}
};
思路3 **- 时间复杂度: O(lg(n+m))*- 空间复杂度: O(1)***
由于题目中建议我们在时间复杂度O(lg(m+n))中完成，我们可以把这题看成寻找第k大的值，这样我们可以递归的去做，每次查找k/2，知道k等于1，注意边界值的处理
class Solution {
public:
int getKth(vector<int> nums1, int start1, int end1, vector<int> nums2, int start2, int end2, int k) {
int len1 = end1 - start1 + 1;
int len2 = end2 - start2 + 1;
if (len1 > len2) return getKth(nums2, start2, end2, nums1, start1, end1, k);
if (len1 == 0) return nums2[start2 + k - 1];
if (k == 1) return min(nums1[start1], nums2[start2]);
int i = start1 + min(len1, k / 2) - 1;
int j = start2 + min(len2, k / 2) - 1;
if (nums1[i] > nums2[j]) {
return getKth(nums1, start1, end1, nums2, j + 1, end2, k - (j - start2 + 1));
}
else {
return getKth(nums1, i + 1, end1, nums2, start2, end2, k - (i - start1 + 1));
}
}
double findMedianSortedArrays(vector<int> nums1, vector<int> nums2) {
int n = nums1.size();
int m = nums2.size();
int left = (n + m + 1) / 2;
int right = (n + m + 2) / 2;
return (getKth(nums1, 0, n - 1, nums2, 0, m - 1, left) + getKth(nums1, 0, n - 1, nums2, 0, m - 1, right)) * 0.5;
}
};
