Median of Two Sorted Arrays

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)).

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  1 class Solution {
  2 public:
  3     /*
  4      * 首先的想法是用额外的一个vector保存排好序的数组,快速排序 o(nlogn)
  5      *
  6      */
  7     double _findMedianSortedArrays(vector<int>& nums1, vector<int>& nums2) {
  8         vector<int>::iterator it;
  9         vector<int> all_nums;
 10
 11         for (it=nums1.begin(); it!=nums1.end(); it++) {
 12             all_nums.push_back(*it);
 13         }
 14
 15         for (it=nums2.begin(); it!=nums2.end(); it++) {
 16             all_nums.push_back(*it);
 17         }
 18
 19         sort(all_nums.begin(), all_nums.end());
 20
 21         int len_all_nums = all_nums.size();
 22         if (len_all_nums % 2 == 1) {
 23             return all_nums[len_all_nums / 2];
 24         } else {
 25             double mid_left = all_nums[len_all_nums / 2 - 1];
 26             double mid_right = all_nums[len_all_nums / 2];
 27             double mid_num = (mid_left + mid_right) / 2;
 28             return mid_num;
 29         }
 30     }
 31     /*
 32      * 也可以转换为找到两个排序链表的第i个数, 不过太麻烦了
 33      * 1.
 34      * */
 35    double findMedianSortedArrays(vector<int>& nums1, vector<int>& nums2) {
 36         int all_len = nums1.size() + nums2.size();
 37         int first_index = -1;
 38         int second_index = -1;
 39         if (all_len % 2 == 1) {
 40             first_index = all_len / 2;
 41         } else {
 42             first_index = all_len / 2 - 1;
 43             second_index = all_len / 2;
 44         }
 45         double first_num = 0;
 46         double second_num = 0;
 47         //下面就是找到对应的数
 48         int n1 = 0;
 49         int n2 = 0;
 50         int find_index = 0;
 51         for (;find_index<first_index; find_index++) {
 52             if ((n1 < nums1.size()) && (n2< nums2.size())) {
 53                 if (nums1[n1] <= nums2[n2]) {
 54                     n1++;
 55                 } else {
 56                     n2++;
 57                 }
 58             } else if(n1 < nums1.size()) {
 59                 n1++;
 60             } else {
 61                 n2++;
 62             }
 63         }
 64
 65         if (n1 == nums1.size()) {
 66             first_num = nums2[n2];
 67             if (second_index != -1) {
 68                 second_num = nums2[n2 + 1];
 69             }
 70         } else if (n2 == nums2.size()) {
 71             first_num = nums1[n1];
 72             if (second_index != -1) {
 73                 second_num = nums1[n1 + 1];
 74             }
 75         } else {
 76             /*
 77             first_num = min(nums1[n1], nums2[n2]);
 78             //second_num = max(nums1[n1], nums2[n2]);
 79             if ((n1 + 1) < nums1.size()) {
 80                 second_num = min(nums1[n1 + 1], nums2[n2]);
 81             } else {
 82                 second_num = nums2[n2];
 83             }
 84             */
 85             if (nums1[n1] <= nums2[n2]) {
 86                 first_num = nums1[n1];
 87                 if ((n1 + 1) < nums1.size()) {
 88                     second_num = min(nums1[n1 + 1], nums2[n2]);
 89                 } else {
 90                     second_num = nums2[n2];
 91                 }
 92             } else {
 93                 first_num = nums2[n2];
 94                 if ((n2 + 1) < nums2.size()) {
 95                     second_num = min(nums1[n1], nums2[n2 + 1]);
 96                 } else {
 97                     second_num = nums1[n1];
 98                 }
 99             }
100         }
101
102         if (second_index != -1) {
103             return (first_num + second_num) / 2.0;
104         } else {
105             return first_num;
106         }
107    }
108 };