feat: 修改错别字，并添加题解 #343
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feat: 修改错别字，并添加题解 #343
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azl397985856
requested changes
Apr 14, 2020
problems/1.TwoSum.md
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| 所以返回 [0, 1] | ||
| ``` | ||
| ## 思路 | ||
| 使用一层循环、每遍历到一个元素就计算该元素与target之间的差值diff，然后以diff为下标到数组temp中寻找 |
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- 题目都加下时间复杂度分析吧。 我后面的题目也会加。 这题的复杂度不好，为O（max（nums）），建议 hashmap
- 可以先将一下双层循环怎么做
problems/1.TwoSum.md
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| - 如果temp[diff]有值，则返回两个元素在数组nums的下标 | ||
| - 如果没有找到，则将当前元素存入数组temp中(下标nums[i], 下标nums[i]对应的值i) | ||
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| 时间复杂度:O(n) |
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复杂度的格式为：
复杂度分析
- 时间复杂度：$O(N)$
- 空间复杂度：$O(N)$
problems/1.TwoSum.md
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| const temp = []; | ||
| for (let i = 0; i < nums.length; i++) { | ||
| const diff = target - nums[i]; | ||
| if (temp[diff] != undefined) { |
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Suggested change
| if (temp[diff] != undefined) { | |
| if (temp[diff] != void 0) { |
| 输出: false | ||
| ``` | ||
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| ## 思路 |
| ``` | ||
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| ## 思路 | ||
| - 用哈希表存下对应的值 |
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| ``` | ||
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| ## 思路 |
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一样的问题。 这不是思路，是关键点。
并且使用递归的原因只是其比较简单，容易书写，并且数据规模合适，并不是关键
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| ``` | ||
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| ## 思路 |
problems/22.GenerateParentheses.md
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| ## 关键点 | ||
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| - 处理有效括号 |
| 所以你应该输出2. | ||
| ``` | ||
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| ## 思路 |
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建议改为从小到大满足。 参考：
class Solution:
def findContentChildren(self, g: List[int], s: List[int]) -> int:
g.sort()
s.sort()
cnt = 0
j = 0
for i in range(len(s)):
if j < len(g) and s[i] >= g[j]:
cnt += 1
j += 1
return cnt| ] | ||
| ``` | ||
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| ## 思路 |
azl397985856
requested changes
Apr 15, 2020
problems/1.TwoSum.md
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| - 最容易想到的就是暴力枚举，我们可以利用两层 for 循环来遍历每个元素，并查找满足条件的目标元素。 | ||
| - 不过这样时间复杂度为 O(N^2)，空间复杂度为 O(1)，时间复杂度较高，我们要想办法进行优化。 | ||
| - 我们增加一个 Map 记录已经遍历过的数字及其对应的索引值。 | ||
| - 这样当遍历一个新数字的时候去 Map 里查询，与该数的差值是否已经在前面的数字中出现过 |
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Suggested change
| - 这样当遍历一个新数字的时候去 Map 里查询，与该数的差值是否已经在前面的数字中出现过 | |
| - 这样当遍历一个新数字的时候去 Map 里查询，target 与该数的差值是否已经在前面的数字中出现过 |
| @@ -15,16 +15,22 @@ https://leetcode-cn.com/problems/two-sum | |||
| 所以返回 [0, 1] | |||
| ``` | |||
| ## 思路 | |||
| ## 思路 | ||
| - 用哈希表存下对应的值 | ||
| - 递归 | ||
| ## 思想 |
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Suggested change
| ## 思想 | |
| ## 思路 |
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| ## 关键点 | ||
| 利用回溯思想解题，在for循环中调用递归。 |
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Suggested change
| 利用回溯思想解题，在for循环中调用递归。 | |
| - 回溯 |
problems/52.N-Queens-II.md
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| ## 关键点 | ||
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| 对位运算公式的理解和运用 |
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Suggested change
| 对位运算公式的理解和运用 | |
| - 位运算 | |
| - DFS（深度优先搜索） |
azl397985856
approved these changes
Apr 15, 2020
azl397985856
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