|
| 1 | +## 每日一题 - Find All Numbers Disappeared in an Array |
| 2 | + |
| 3 | +### 信息卡片 |
| 4 | + |
| 5 | +- 时间:2019-06-05 |
| 6 | +- 题目链接:https://leetcode.com/problems/find-all-numbers-disappeared-in-an-array/ |
| 7 | +- tag:`array` |
| 8 | + |
| 9 | +### 题目描述 |
| 10 | + |
| 11 | +``` |
| 12 | +Given an array of integers where 1 ≤ a[i] ≤ n (n = size of array), some elements appear twice and others appear once. |
| 13 | +
|
| 14 | +Find all the elements of [1, n] inclusive that do not appear in this array. |
| 15 | +
|
| 16 | +Could you do it without extra space and in O(n) runtime? You may assume the returned list does not count as extra space. |
| 17 | +
|
| 18 | +Example: |
| 19 | +
|
| 20 | +Input: |
| 21 | +[4,3,2,7,8,2,3,1] |
| 22 | +
|
| 23 | +Output: |
| 24 | +[5,6] |
| 25 | +``` |
| 26 | + |
| 27 | +### 参考答案 |
| 28 | + |
| 29 | +#### 使用额外的空间记录出现过的数字, 时间复杂度和空间复杂度皆为O(n) |
| 30 | + |
| 31 | +参考代码 |
| 32 | +```javascript |
| 33 | +/* |
| 34 | + * @lc app=leetcode id=448 lang=javascript |
| 35 | + * |
| 36 | + * [448] Find All Numbers Disappeared in an Array |
| 37 | + */ |
| 38 | +/** |
| 39 | + * @param {number[]} nums |
| 40 | + * @return {number[]} |
| 41 | + */ |
| 42 | +var findDisappearedNumbers = function(nums) { |
| 43 | + let allNums = []; |
| 44 | + let res = []; |
| 45 | + |
| 46 | + for (let i = 0; i < nums.length; i++){ |
| 47 | + allNums[nums[i] - 1] = true; |
| 48 | + } |
| 49 | + |
| 50 | + for (let i = 0; i < nums.length; i++){ |
| 51 | + if(!allNums[i]){ |
| 52 | + res.push(i + 1); |
| 53 | + } |
| 54 | + } |
| 55 | + return res; |
| 56 | +}; |
| 57 | +``` |
| 58 | + |
| 59 | +#### 充分利用题目 "You may assume the returned list does not count as extra space." |
| 60 | + |
| 61 | +- 用res记录哪些数字出现过 |
| 62 | +- 最后遍历时, 判断res是否为空, 若是, 则证明未出现过, 将其写回res |
| 63 | + |
| 64 | +参考代码 |
| 65 | +```javascript |
| 66 | +/* |
| 67 | + * @lc app=leetcode id=448 lang=javascript |
| 68 | + * |
| 69 | + * [448] Find All Numbers Disappeared in an Array |
| 70 | + */ |
| 71 | +/** |
| 72 | + * @param {number[]} nums |
| 73 | + * @return {number[]} |
| 74 | + */ |
| 75 | +var findDisappearedNumbers = function(nums) { |
| 76 | + const res = []; |
| 77 | + let cur = 0; |
| 78 | + for(let i = 0; i < nums.length; i++) { |
| 79 | + res[nums[i]] = true; |
| 80 | + } |
| 81 | + |
| 82 | + for(let i = 0; i < nums.length; i++) { |
| 83 | + if (res[i + 1] === void 0) { |
| 84 | + res[cur++] = i + 1; |
| 85 | + } |
| 86 | + } |
| 87 | + |
| 88 | + res.length = cur; |
| 89 | + |
| 90 | + return res; |
| 91 | +}; |
| 92 | +``` |
| 93 | + |
| 94 | + |
| 95 | +### 其他优秀解答 |
| 96 | +利用python集合类型的特点: 元素唯一, 不存在相同元素 |
| 97 | + |
| 98 | +```python |
| 99 | +class Solution(object): |
| 100 | + def findDisappearedNumbers(self, nums): |
| 101 | + """ |
| 102 | + :type nums: List[int] |
| 103 | + :rtype: List[int] |
| 104 | + """ |
| 105 | + ls = [i for i in range(1, len(nums)+1)] |
| 106 | + |
| 107 | + return list(set(ls) - set(nums)) |
| 108 | +``` |
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