|
| 1 | +--- |
| 2 | +title: leetcode4 寻找两个正序数组的中位数【困难难度】 |
| 3 | +tags: |
| 4 | + - leetcode |
| 5 | + - 学习笔记 |
| 6 | + - 算法 |
| 7 | +--- |
| 8 | + |
| 9 | +### [4. 寻找两个正序数组的中位数](https://leetcode-cn.com/problems/median-of-two-sorted-arrays/) |
| 10 | + |
| 11 | +### 英文题目: Median of two sorted arrays |
| 12 | + |
| 13 | +<table> <tr> <td bgcolor=white width=auto> ●くろまる 难度: </td> <td bgcolor=#D9534F width=auto><font color=white>困难</font></td> </tr></table> |
| 14 | + |
| 15 | +给定两个大小分别为 `m` 和 `n` 的正序(从小到大)数组 `nums1` 和 `nums2`。请你找出并返回这两个正序数组的 **中位数** 。 |
| 16 | + |
| 17 | + |
| 18 | + |
| 19 | +**示例 1:** |
| 20 | + |
| 21 | +``` |
| 22 | +输入:nums1 = [1,3], nums2 = [2] |
| 23 | +输出:2.00000 |
| 24 | +解释:合并数组 = [1,2,3] ,中位数 2 |
| 25 | +``` |
| 26 | + |
| 27 | +**示例 2:** |
| 28 | + |
| 29 | +``` |
| 30 | +输入:nums1 = [1,2], nums2 = [3,4] |
| 31 | +输出:2.50000 |
| 32 | +解释:合并数组 = [1,2,3,4] ,中位数 (2 + 3) / 2 = 2.5 |
| 33 | +``` |
| 34 | + |
| 35 | +**示例 3:** |
| 36 | + |
| 37 | +``` |
| 38 | +输入:nums1 = [0,0], nums2 = [0,0] |
| 39 | +输出:0.00000 |
| 40 | +``` |
| 41 | + |
| 42 | +**示例 4:** |
| 43 | + |
| 44 | +``` |
| 45 | +输入:nums1 = [], nums2 = [1] |
| 46 | +输出:1.00000 |
| 47 | +``` |
| 48 | + |
| 49 | +**示例 5:** |
| 50 | + |
| 51 | +``` |
| 52 | +输入:nums1 = [2], nums2 = [] |
| 53 | +输出:2.00000 |
| 54 | +``` |
| 55 | + |
| 56 | + |
| 57 | + |
| 58 | +**提示:** |
| 59 | + |
| 60 | +- `nums1.length == m` |
| 61 | +- `nums2.length == n` |
| 62 | +- `0 <= m <= 1000` |
| 63 | +- `0 <= n <= 1000` |
| 64 | +- `1 <= m + n <= 2000` |
| 65 | +- `-106 <= nums1[i], nums2[i] <= 106` |
| 66 | + |
| 67 | + |
| 68 | +**进阶:**你能设计一个时间复杂度为 `O(log (m+n))` 的算法解决此问题吗? |
| 69 | + |
| 70 | +<br/> |
| 71 | + |
| 72 | +### 思路 |
| 73 | +先合并两个有序数组,然后根据数组长度的奇偶来取到中位数。如果是偶数个,就取中间两个的平均数;如果是奇数个,直接取最中间的即可。 |
| 74 | + |
| 75 | +与 `leetcode 88. 合并两个有序数组` 类似。 |
| 76 | + |
| 77 | +<br/> |
| 78 | + |
| 79 | + |
| 80 | +### 已AC的C++代码: |
| 81 | +```cpp |
| 82 | +class Solution { |
| 83 | +public: |
| 84 | + double findMedianSortedArrays(vector<int>& nums1, vector<int>& nums2) { |
| 85 | + vector<int> nums; |
| 86 | + double res; |
| 87 | + int m = nums1.size(); |
| 88 | + int n = nums2.size(); |
| 89 | + int len; |
| 90 | + |
| 91 | + int i = 0, j = 0; |
| 92 | + while (i < m && j < n) // 只要一个指针扫到数组末尾,循环结束 |
| 93 | + { |
| 94 | + if(nums1[i] <= nums2[j]) |
| 95 | + { |
| 96 | + nums.push_back(nums1[i]); |
| 97 | + i++; |
| 98 | + } |
| 99 | + else { |
| 100 | + nums.push_back(nums2[j]); |
| 101 | + j++; |
| 102 | + } |
| 103 | + } |
| 104 | + |
| 105 | + while(i < m) // 数组nums1没跑完,nums2已跑完时 |
| 106 | + { |
| 107 | + nums.push_back(nums1[i]); |
| 108 | + i++; |
| 109 | + } |
| 110 | + |
| 111 | + while(j < n) // 数组nums2没跑完,nums1已跑完时 |
| 112 | + { |
| 113 | + nums.push_back(nums2[j]); |
| 114 | + j++; |
| 115 | + } |
| 116 | + len = nums.size(); |
| 117 | + if(len % 2 == 0) |
| 118 | + { |
| 119 | + res = (nums[len/2] + nums[len/2-1])/2.0; |
| 120 | + } |
| 121 | + else res = nums[len/2]; |
| 122 | + |
| 123 | + return res; |
| 124 | + } |
| 125 | +}; |
| 126 | +``` |
| 127 | + |
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