Median of Two Sorted Arrays
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题目: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)).
思路:
本题如此讲解,设两个数组:a[0...i....m] , b[0...j....n],两数组排列完毕。
如果a[i]<b[j] , 那么对a数组从i开始迭代寻找,前面那段没必要寻找。
循环递归结束的条件是k=1的时候,返回第一个数字。
代码:
class Solution {public: double findMedianSortedArrays(vector<int>& nums1, vector<int>& nums2) { int len=nums1.size()+nums2.size(); if(len%2)//奇数 return findKthElement( nums1 ,nums2 , (len+1)/2 ) ; else//偶数 return (findKthElement(nums1,nums2,len/2)+findKthElement( nums1 , nums2 , len/2+1 ))*0.5 ; } double min(double x,double y){ return x<y?x:y; } double findKthElement(vector<int>& nums1,vector<int>& nums2,int k){ if(nums1.size()>nums2.size()){ return findKthElement(nums2,nums1,k); } if(nums1.empty()) return nums2[k-1]; if(nums2.empty()) return nums1[k-1]; if(k==1) return min(nums1[0],nums2[0]); int ia=min(k/2,nums1.size()); int ib=k-ia; if(nums1[ia-1]<nums2[ib-1]){ vector<int> new_nums1(nums1.begin()+ia,nums1.end()); return findKthElement( new_nums1 , nums2 , k-ia ) ; }if(nums1[ia-1]>nums2[ib-1]){ vector<int> new_nums2(nums2.begin()+ib,nums2.end()); return findKthElement( nums1 , new_nums2 , k-ib ) ; } return nums1[ia-1]; }};
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