算法分析与设计——LeetCode Problem.4 Median of Two Sorted Arrays

题目链接

问题详情

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

解题思路

比较容易想到的方法是将两个数组合并排序后找出中位数，但是这样做的复杂度是O(n)，但题目要求的复杂度是O(log n)。然而我不理解的是这个方法的代码依然
通过了评测。代码如下：

class Solution {public:double findMedianSortedArrays(vector<int>& nums1, vector<int>& nums2) {vector<int> v;int p1 = 0, p2 = 0;while (p1 < nums1.size() && p2 < nums2.size()) {if (nums1[p1] < nums2[p2]) {v.push_back(nums1[p1]);p1++;}else {v.push_back(nums2[p2]);p2++;}}if (p1 == nums1.size()) {while (p2 < nums2.size()) {v.push_back(nums2[p2]);p2++;}}if (p2 == nums2.size()) {while (p1 < nums1.size()) {v.push_back(nums1[p1]);p1++;}}if (v.size() % 2 == 1) {return double(v[v.size() / 2]);}else {return double((v[v.size() / 2] + v[v.size() / 2 - 1] / 2.0));}}};

然后我从网上学习了更好的做法。

可以假设中位数是两个数组中第k小的数字，然后不断去掉两个数组中肯定不是中位数的部分，此时k的值随之变化，最后当k的值为1或者有一数组为空时即可找到
中位数。代码如下：

class Solution {public:    double findMedian(vector<int> v1, vector<int> v2, int median) {    //确保v1长度比v2小    if (v1.size() > v2.size()) {    swap(v1, v2);    }    if (v1.size() == 0) {    return v2[median - 1];    }    int i1 = v1.size();    int i2 = v2.size();    if (median == 1) {//取两个数组中的最小值if (v1[0] < v2[0]) {return v1[0];}return v2[0];    }    int t1 = min(i1, median / 2);    int t2 = min(i2, median / 2);    //    if (v1[t1-1] > v2[t2-1]) {    return findMedian(v1, vector<int> (v2.begin()+t2, v2.end()), median - t2);    }    else {    return findMedian(vector<int> (v1.begin()+t1, v1.end()), v2, median - t1);    }    return 0;    }    double findMedianSortedArrays(vector<int>& nums1, vector<int>& nums2) {    //当nums1或nums2为空    if (nums2.size() == 0) {    if (nums1.size() % 2 == 1) {    return nums1[nums1.size() / 2];    }    else {    return double((nums1[nums1.size() / 2] + nums1[nums1.size() / 2 - 1]) / 2.0);    }    }    if (nums1.size() == 0) {    if (nums2.size() % 2 == 1) {    return nums2[nums2.size() / 2];    }    else {    return double((nums2[nums2.size() / 2] + nums2[nums2.size() / 2 - 1]) / 2.0);    }    }    int n1 = nums1.size(), n2 = nums2.size();    if ((n1 + n2) % 2 == 1) {    return findMedian(nums1, nums2, (n1 + n2) / 2.0 + 1);    }    else {    return (findMedian(nums1, nums2, (n1 + n2) / 2.0) + findMedian(nums1, nums2, (n1 + n2) / 2 + 1)) / 2.0;    }    }};