假设我们有两个数组;这些数组已排序。因此,我们必须找到这两个数组的中位数。因此,如果数组类似于[1,5,8]和[2,3,6,9],则答案将为5。
为了解决这个问题,我们将遵循以下步骤-
定义一个函数findMedianSortedArrays,它将使用nums1和nums2数组
如果nums1的大小> nums2的大小,那么,
调用函数return findMedianSortedArrays(nums2,nums1)
x:= nums1的大小,y:= nums2的大小
低:= 0,高:= x
totalLength:= x + y
当低<=高时,执行-
高:= partitionX-1
如果totalLength mod 2与0相同,则
除此以外
return(maxLeftX和maxLeftY的最大值)+ minRightX和minRightY的最小值)/ 2
返回maxLeftX和maxLeftY的最大值
partitionX:=低+(高-低)/ 2
partitionY:=(totalLength + 1)/ 2-partitionX
当partitionX为0时,maxLeftX = -inf,否则为nums1 [partitionX-1]
当partitionX为x时,maxRightX = inf,否则为nums1 [partitionX]
当partitionY为0时,maxLeftY = -inf,否则为nums2 [partitionY-1]
当partitionY为y时,maxRightY = inf,否则为nums2 [partitionY]
如果maxLeftX <= minRightY并且maxLeftY <= minRightX,则,
否则,当maxLeftX> minRightY时,则-
否则低:= partitionX + 1
返回0
让我们看下面的实现以更好地理解-
#include <bits/stdc++.h>
using namespace std;
class Solution {
public:
double findMedianSortedArrays(vector& nums1, vector<int>& nums2) {
if(nums1.size()>nums2.size())
return findMedianSortedArrays(nums2,nums1);
int x = nums1.size();
int y = nums2.size();
int low = 0;
int high = x;
int totalLength = x+y;
while(low<=high){
int partitionX = low + (high - low)/2;
int partitionY = (totalLength + 1)/2 - partitionX;
int maxLeftX = (partitionX ==0?INT_MIN:nums1[partitionX1] );
int minRightX = (partitionX == x?INT_MAX :
nums1[partitionX]);
int maxLeftY = (partitionY ==0?INT_MIN:nums2[partitionY1] );
int minRightY = (partitionY == y?INT_MAX : nums2[partitionY]);
if(maxLeftX<=minRightY && maxLeftY <= minRightX){
if(totalLength% 2 == 0){
return ((double)max(maxLeftX,maxLeftY) + (double)min(minRightX,minRightY))/2;
} else {
return max(maxLeftX, maxLeftY);
}
}
else if(maxLeftX>minRightY)
high = partitionX-1;
else low = partitionX+1;
}
return 0;
}
};
main(){
Solution ob;
vector<int> v1 = {1,5,8}, v2 = {2,3,6,9};
cout << (ob.findMedianSortedArrays(v1, v2));
}[1,5,8] [2,3,6,9]
输出结果
5