LeetCode 4. Median of Two Sorted Arrays

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)).

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：
就是求两个数组合并后的数组的中位数，利用归并思想合并，再求中位数，不难。
计算nums1和nums2的长度len1、len2，记len为len1和len2的和，如果和都0，返回double类型的0。利用归并思想，如果i小于len1,j小于len2，将nums1[i]和nums2[j]中小的那个值放入res，并对应索引加1。上面循环结束，如果是i

class Solution {
public:double findMedianSortedArrays(vector<int>& nums1, vector<int>& nums2) {int len1=nums1.size();//计算nums1和nums2的长度int len2=nums2.size();int len=len1+len2;//记len为len1和len2的和if(len==0){//如果和都0，返回double类型的0return double(0);}vector<int> res;int i=0,j=0;//利用归并思想while(i<len1&&j<len2){//如果i小于len1,j小于len2res.push_back(nums1[i]<nums2[j]?nums1[i++]:nums2[j++]);//将nums1[i]和nums2[j]中小的那个值放入res，并对应索引加1}while(i<len1){//上面循环结束，如果是i<len1，也就是j已经到达len2，将nums1中剩余的数放入resres.push_back(nums1[i++]);}while(j<len2){//上面循环结束，如果是j<len2，也就是i已经到达len1，将nums2中剩余的数放入resres.push_back(nums2[j++]);}if((len1+len2)%2==0){//如果总个数为偶数个，则中位数为索引值为len/2与len/2-1的两个值转换为double类型后的平均值return ((double(res[len/2])+double(res[len/2-1]))/2);}else{//如果总个数为奇数，中位数为索引值为len/2的那个值转换为double类型后的值return double(res[len/2]);}}
};

输出结果： 52ms

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