From a72cff447c679eb399e1e023cee2bbfe96e06a09 Mon Sep 17 00:00:00 2001 From: fanxb Date: Thu, 10 Mar 2022 17:23:59 +0800 Subject: [PATCH] add --- 算法/leetcode/常规题/Q131 分割回文串.md | 4 +- 算法/leetcode/常规题/Q216.组合总和 3.md | 68 ++++++++++++++++++ 算法/leetcode/常规题/Q39.组合总和.md | 69 +++++++++++++++++++ 算法/leetcode/常规题/Q40.组合总和 2.md | 68 ++++++++++++++++++ .../leetcode/常规题/xml/q131-字符串切割.drawio | 1 + 算法/leetcode/常规题/xml/q39.drawio | 1 + .../leetcode/面试题/xml/q131-字符串切割.drawio | 1 - 7 files changed, 209 insertions(+), 3 deletions(-) create mode 100644 算法/leetcode/常规题/Q216.组合总和 3.md create mode 100644 算法/leetcode/常规题/Q39.组合总和.md create mode 100644 算法/leetcode/常规题/Q40.组合总和 2.md create mode 100644 算法/leetcode/常规题/xml/q131-字符串切割.drawio create mode 100644 算法/leetcode/常规题/xml/q39.drawio delete mode 100644 算法/leetcode/面试题/xml/q131-字符串切割.drawio diff --git a/算法/leetcode/常规题/Q131 分割回文串.md b/算法/leetcode/常规题/Q131 分割回文串.md index 10be163..858ac2c 100644 --- a/算法/leetcode/常规题/Q131 分割回文串.md +++ b/算法/leetcode/常规题/Q131 分割回文串.md @@ -2,7 +2,7 @@ id: "20220309" date: "2022/03/09 20:15" title: "Q131 分割回文串(palindrome partitioning)" -tags: ["java", "leetcode", "dp", "dfs"] +tags: ["java", "leetcode", "dp", "dfs","回溯"] index_img: https://qiniupic.fleyx.com/blog/202203091650691.jpg?imageView2/2/w/200 banner_img: https://qiniupic.fleyx.com/blog/202203091650691.jpg categories: @@ -20,7 +20,7 @@ leetcode 中等难度，题目描述[点击这里](https://leetcode-cn.com/probl 判断回文数的方法很简单，两个指针从两端往中间遍历即可，或者先用 dp 计算得到所有的回文数组合（本解答采用 dp）,本题字符串最长为 16，暴力判断和 dp 区别不大。 -关键是如何穷举出所有分割可能，简单的 for 循环显然是做不到的.具体分割方法可以先看下图： +关键是如何穷举出所有分割可能，简单的 for 循环显然是做不到的.需要用到回溯思想，具体分割方法可以先看下图： ## 字符串分割 diff --git a/算法/leetcode/常规题/Q216.组合总和 3.md b/算法/leetcode/常规题/Q216.组合总和 3.md new file mode 100644 index 0000000..18ad2e7 --- /dev/null +++ b/算法/leetcode/常规题/Q216.组合总和 3.md @@ -0,0 +1,68 @@ +--- +id: "202203103" +date: "2022/03/10 22:15" +title: "Q216 组合总和3(combination-sumⅢ)" +tags: ["java", "leetcode", "dfs"] +index_img: https://qiniupic.fleyx.com/blog/202203101631050.jpg?imageView2/2/w/200 +banner_img: https://qiniupic.fleyx.com/blog/202203101631050.jpg +categories: + - "算法" + - "leetcode刷题" +--- + + + +## 解析思路 + +leetcode 中等难度，题目描述[点击这里](https://leetcode-cn.com/problems/combination-sum-iii/)。 + +本题属于[组合排序 2](https://blog.fleyx.com/blog/detail/202203102/)的进阶题型。建议先看[上一篇](https://blog.fleyx.com/blog/detail/202203102/)的解析。 + +本题跟上一题区别只有一点是**选择的数字不能重复**,**同时对选择的数量有限制**如何实现呢？ + +很简单，只需要 + +- 将传入的 index 设置成 index+1，实现不重复 +- ~~同时 for 循环时每次都循环到下一个不同的元素上~~由于本来提供的数字就不重复，因此可以去掉这一条 +- 循环和返回结果是都判断下元素数量 + +代码如下： + +## 代码 + +```java +public class Q216 { + public List> combinationSum3(int k, int n) { + int[] arr = {1, 2, 3, 4, 5, 6, 7, 8, 9}; + List> res = new ArrayList<>(); + dfs(arr, n, k, 0, new Stack<>(), res); + return res; + } + + private void dfs(int[] candidates, int n, int k, int index, Stack temp, List> res) { + if (n == 0) { + //说明找到一个结果序列 + if (temp.size() == k) { + //只有刚好k个数才是结果 + res.add(new ArrayList<>(temp)); + } + return; + } + for (int i = index; i < candidates.length; i++) { + if (candidates[i] > n || temp.size() == k) { + //前面已经排序过，所以在这里可以进行剪枝操作，如果candidates[index]都小于target了，那就不需要比较后面的了，肯定不满足要求 + //另外只能使用k个数，所以当temp的size为k时说明不能在装了，本条路结束 + return; + } + temp.push(candidates[i]); + dfs(candidates, n - candidates[i], k, i + 1, temp, res); + temp.pop(); + } + } + + public static void main(String[] args) { + new Q216().combinationSum3(3, 9).forEach(System.out::println); + } + +} +``` diff --git a/算法/leetcode/常规题/Q39.组合总和.md b/算法/leetcode/常规题/Q39.组合总和.md new file mode 100644 index 0000000..2ec0942 --- /dev/null +++ b/算法/leetcode/常规题/Q39.组合总和.md @@ -0,0 +1,69 @@ +--- +id: "202203101" +date: "2022/03/10 20:15" +title: "Q39 组合总和(combination-sum)" +tags: ["java", "leetcode", "dfs"] +index_img: https://qiniupic.fleyx.com/blog/202203101631050.jpg?imageView2/2/w/200 +banner_img: https://qiniupic.fleyx.com/blog/202203101631050.jpg +categories: + - "算法" + - "leetcode刷题" +--- + + + +## 解析思路 + +leetcode 中等难度，题目描述[点击这里](https://leetcode-cn.com/problems/combination-sum/)。 + +标准的回溯类题目，对于回溯类题目，通常都是要穷举所有的情况，然后判断那些情况是符合题目要求的。 + +然后穷举通常是通过`深度优先搜索(dfs)`来实现的，我们可以先将结果展开成一棵树，然后再根据这棵树来写代码，就比较好理解，如下图： + +以[2,3,5],8 为例： + +![展开](https://qiniupic.fleyx.com/blog/202203101700108.png) + +上图只对最左边的路径进行了完全展开（全部展开太麻烦了） + +总的来说回溯就是不断的进行尝试，如果尝试到最后发现不满足要求那就换一个路径继续尝试，属于暴力算法。因此此类题目的数据规模通常会限制的比较小。 + +另外通常还能根据题目的要求来做剪枝操作，减少一些不必要的运算。 + +比如本题可以先将数组排序，当选择的某个数已经大于目标值时，就没必须选择这个数的下一个数继续尝试了。 + +## 代码 + +```java +public class Q39 { + public List> combinationSum(int[] candidates, int target) { + List> res = new ArrayList<>(); + //排序 + Arrays.sort(candidates); + dfs(candidates, target, 0, new Stack<>(), res); + return res; + } + + private void dfs(int[] candidates, int target, int index, Stack temp, List> res) { + if (target == 0) { + //说明找到一个结果序列 + res.add(new ArrayList<>(temp)); + return; + } + for (int i = index; i < candidates.length; i++) { + if (candidates[i] > target) { + //前面已经排序过，所以在这里可以进行剪枝操作，如果candidates[index]都小于target了，那就不需要比较后面的了，肯定不满足要求 + return; + } + temp.push(candidates[i]); + //注意这里，为了让结果集不重复，选择重复元素时只能对当前元素进行重复选择，不能重复选择之前的元素。所以递归的index为i,不是0 + dfs(candidates, target - candidates[i], i, temp, res); + temp.pop(); + } + } + + public static void main(String[] args) { + new Q39().combinationSum(new int[]{2, 3, 5}, 8).forEach(System.out::println); + } +} +``` diff --git a/算法/leetcode/常规题/Q40.组合总和 2.md b/算法/leetcode/常规题/Q40.组合总和 2.md new file mode 100644 index 0000000..ac16a70 --- /dev/null +++ b/算法/leetcode/常规题/Q40.组合总和 2.md @@ -0,0 +1,68 @@ +--- +id: "202203102" +date: "2022/03/10 21:15" +title: "Q40 组合总和2(combination-sumⅡ)" +tags: ["java", "leetcode", "dfs"] +index_img: https://qiniupic.fleyx.com/blog/202203101631050.jpg?imageView2/2/w/200 +banner_img: https://qiniupic.fleyx.com/blog/202203101631050.jpg +categories: + - "算法" + - "leetcode刷题" +--- + + + +## 解析思路 + +leetcode 中等难度，题目描述[点击这里](https://leetcode-cn.com/problems/combination-sum-ii/)。 + +本题属于[组合排序](https://blog.fleyx.com/blog/detail/202203101/)的进阶题型。建议先看[上一篇](https://blog.fleyx.com/blog/detail/202203101/)的解析。 + +本题跟上一题区别只有一点是**选择的数字不能重复**,如何实现不重复呢？ + +很简单，只需要将传入的 index 设置成 index+1，同时 for 循环时每次都循环到下一个不同的元素上 + +这样就不会选取到已经选过的元素。 + +代码如下： + +## 代码 + +```java +public class Q40 { + public List> combinationSum2(int[] candidates, int target) { + List> res = new ArrayList<>(); + Arrays.sort(candidates); + dfs(candidates, target, 0, new Stack<>(), res); + return res; + } + + private void dfs(int[] candidates, int target, int index, Stack temp, List> res) { + if (target == 0) { + //说明找到一个结果序列 + res.add(new ArrayList<>(temp)); + return; + } + for (int i = index; i < candidates.length; ) { + if (candidates[i] > target) { + //前面已经排序过，所以在这里可以进行剪枝操作，如果candidates[index]都小于target了，那就不需要比较后面的了，肯定不满足要求 + return; + } + temp.push(candidates[i]); + dfs(candidates, target - candidates[i], i + 1, temp, res); + temp.pop(); + //手动控制i的增长，对于同一个数字不能重复处理 + int nextI = i + 1; + while (nextI < candidates.length && candidates[nextI] == candidates[i]) { + nextI++; + } + i = nextI; + } + } + + public static void main(String[] args) { + new Q40().combinationSum2(new int[]{10, 1, 2, 7, 6, 1, 5}, 8).forEach(System.out::println); + } + +} +``` diff --git a/算法/leetcode/常规题/xml/q131-字符串切割.drawio b/算法/leetcode/常规题/xml/q131-字符串切割.drawio new file mode 100644 index 0000000..9560f59 --- /dev/null +++ b/算法/leetcode/常规题/xml/q131-字符串切割.drawio @@ -0,0 +1 @@ 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\ No newline at end of file