LeetCode练习_20170519

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

思路：把两个数组合并为一个，然后根据位置直接得到对应元素并计算结果（因为都是有序数组，依次比较最前面的即可）

double findMedianSortedArrays(vector<int>& nums1, vector<int>& nums2) {        int n1 = nums1.size();        int n2 = nums2.size();        vector<int> nums3;        vector<int>::iterator it1 = nums1.begin();        vector<int>::iterator it2 = nums2.begin();                while(!(it1 == nums1.end() && it2 == nums2.end())){            if(it1 == nums1.end()){                while(it2 != nums2.end()){                    nums3.push_back(*it2);                    it2++;                }            }            else if(it2 == nums2.end()){                while(it1 != nums1.end()){                    nums3.push_back(*it1);                    it1++;                }                            }            else{                if(*it1<=*it2){                    nums3.push_back(*it1);                    it1++;                                    }else{                    nums3.push_back(*it2);                    it2++;                                     }            }        }        //vector<int>::iterator it3 = nums3.begin();        //for(;it3!=nums3.end();it3++){        //    cout<<*it3<<" ";        //}        if(nums3.size()%2 == 0){            float mid = nums3.size()/2 - 1;            return (((float)(*(nums3.begin()+mid))+(float)(*(nums3.begin()+mid+1)))/2);        }else{            float mid = nums3.size()/2;            return (float)(*(nums3.begin()+mid));        }    }