solution:
这个题目想一口气写好不容易。 需要考虑的细节很多。算法倒是很好理解。既然是sorted array, 那么必须用好排好的序。所以要追求 O(ln n).
代码:
class Solution { public: double findMedianSortedArrays(vector<int>& nums1, vector<int>& nums2) { double res; int l1=nums1.size(); int l2=nums2.size(); int total = l1 + l2; if(total & 0x1){ // total is odd; res = findKth(nums1, 0, l1, nums2, 0, l2, total/2 + 1)*1.0; } else { res = (findKth(nums1, 0, l1, nums2, 0, l2, total/2) + findKth(nums1, 0, l1, nums2, 0, l2, total/2 + 1))/2.0; } return res; } private: int findKth(vector<int>& v1, int a1, int b1, vector<int>& v2, int a2, int b2, int k){ int x1=b1-a1; int x2=b2-a2; if(x1>x2) return findKth(v2, a2, b2, v1, a1, b1, k); if(x1==0) return v2[a2+k-1]; if(k==1) return min(v1[a1], v2[a2]); int m=min(x1, k/2); int n=k-m; if(v1[a1+m-1]<v2[a2+n-1]){ return findKth(v1, a1+m, b1, v2, a2, b2, k-m); } else if(v1[a1+m-1]>v2[a2+n-1]){ return findKth(v1, a1, b1, v2, a2+n, b2, k-n); } else { return v1[a1+m-1]; } } };