{"id":724,"date":"2017-11-03T20:48:26","date_gmt":"2017-11-04T03:48:26","guid":{"rendered":"http:\/\/zxi.mytechroad.com\/blog\/?p=724"},"modified":"2018-04-19T08:39:46","modified_gmt":"2018-04-19T15:39:46","slug":"leetcode-216-combination-sum-iii","status":"publish","type":"post","link":"https:\/\/zxi.mytechroad.com\/blog\/searching\/leetcode-216-combination-sum-iii\/","title":{"rendered":"\u82b1\u82b1\u9171 LeetCode 216. Combination Sum III"},"content":{"rendered":"<div>\n<p><iframe loading=\"lazy\" title=\"\u82b1\u82b1\u9171 LeetCode 216. Combination Sum III - \u5237\u9898\u627e\u5de5\u4f5c EP100\" width=\"500\" height=\"375\" src=\"https:\/\/www.youtube.com\/embed\/UwdX19UvoCI?feature=oembed\" frameborder=\"0\" allow=\"accelerometer; autoplay; clipboard-write; encrypted-media; gyroscope; picture-in-picture; web-share\" referrerpolicy=\"strict-origin-when-cross-origin\" allowfullscreen><\/iframe><\/p>\n<p>\u9898\u76ee\u5927\u610f\uff1a\u8f93\u51fa\u6240\u6709\u7528k\u4e2a\u6570\u7684\u548c\u4e3an\u7684\u7ec4\u5408\u3002\u53ef\u4ee5\u4f7f\u7528\u7684\u5143\u7d20\u662f1\u52309\u3002<\/p>\n<p><strong>Problem:<\/strong><\/p>\n<p>Find all possible combinations of\u00a0<i><b>k<\/b><\/i>\u00a0numbers that add up to a number\u00a0<i><b>n<\/b><\/i>, given that only numbers from 1 to 9 can be used and each combination should be a unique set of numbers.<\/p>\n<\/div>\n<div>\n<p><i><b>Example 1:<\/b><\/i><\/p>\n<p>Input:\u00a0<i><b>k<\/b><\/i>\u00a0= 3,\u00a0<i><b>n<\/b><\/i>\u00a0= 7<\/p>\n<p>Output:<\/p>\n<pre class=\"\">[[1,2,4]]\r\n<\/pre>\n<p><i><b>Example 2:<\/b><\/i><\/p>\n<p>Input:\u00a0<i><b>k<\/b><\/i>\u00a0= 3,\u00a0<i><b>n<\/b><\/i>\u00a0= 9<\/p>\n<p>Output:<\/p>\n<pre class=\"\">[[1,2,6], [1,3,5], [2,3,4]]<\/pre>\n<\/div>\n<p>&nbsp;<\/p>\n<p><script async src=\"\/\/pagead2.googlesyndication.com\/pagead\/js\/adsbygoogle.js\"><\/script><br \/>\n<ins class=\"adsbygoogle\" style=\"display: block; text-align: center;\" data-ad-layout=\"in-article\" data-ad-format=\"fluid\" data-ad-client=\"ca-pub-2404451723245401\" data-ad-slot=\"7983117522\"><\/ins><br \/>\n<script>\n     (adsbygoogle = window.adsbygoogle || []).push({});\n<\/script><\/p>\n<p><strong>Idea:<\/strong><\/p>\n<p>DFS + backtracking<\/p>\n<p>bit<\/p>\n<p><a href=\"http:\/\/zxi.mytechroad.com\/blog\/wp-content\/uploads\/2017\/11\/216-ep100.png\"><img loading=\"lazy\" decoding=\"async\" class=\"alignnone size-full wp-image-736\" src=\"http:\/\/zxi.mytechroad.com\/blog\/wp-content\/uploads\/2017\/11\/216-ep100.png\" alt=\"\" width=\"960\" height=\"540\" srcset=\"https:\/\/zxi.mytechroad.com\/blog\/wp-content\/uploads\/2017\/11\/216-ep100.png 960w, https:\/\/zxi.mytechroad.com\/blog\/wp-content\/uploads\/2017\/11\/216-ep100-300x169.png 300w, https:\/\/zxi.mytechroad.com\/blog\/wp-content\/uploads\/2017\/11\/216-ep100-768x432.png 768w, https:\/\/zxi.mytechroad.com\/blog\/wp-content\/uploads\/2017\/11\/216-ep100-624x351.png 624w\" sizes=\"auto, (max-width: 960px) 100vw, 960px\" \/><\/a><\/p>\n<p><strong>Solution:<\/strong><\/p>\n<p>C++<\/p>\n<pre class=\"lang:c++ decode:true \">\/\/ Author: Huahua\r\n\/\/ Runtime: 0 ms\r\nclass Solution {\r\npublic:\r\n    vector&lt;vector&lt;int&gt;&gt; combinationSum3(int k, int n) {\r\n        vector&lt;vector&lt;int&gt;&gt; ans;\r\n        vector&lt;int&gt; cur;\r\n        dfs(k, n, 1, cur, ans);\r\n        return ans;\r\n    }\r\nprivate:\r\n    \/\/ Use k numbers (&gt;= s) to sum up to n\r\n    void dfs(int k, int n, int s, \r\n             vector&lt;int&gt;&amp; cur, vector&lt;vector&lt;int&gt;&gt;&amp; ans) {\r\n        if (k == 0) {\r\n            if (n == 0) ans.push_back(cur);\r\n            return;\r\n        }\r\n        \r\n        for (int i = s; i &lt;= 9; ++i) {\r\n            if (i &gt; n) return;\r\n            cur.push_back(i);\r\n            dfs(k - 1, n - i, i + 1, cur, ans);\r\n            cur.pop_back();\r\n        }\r\n    }\r\n};<\/pre>\n<p>&nbsp;<\/p>\n<p>C++ \/ binary<\/p>\n<pre class=\"lang:c++ decode:true \">class Solution {\r\npublic:\r\n    vector&lt;vector&lt;int&gt;&gt; combinationSum3(int k, int n) {\r\n        vector&lt;vector&lt;int&gt;&gt; ans;\r\n        \r\n        \/\/ 2^9, generate all combinations of [1 .. 9]\r\n        for (int i = 0; i &lt; (1 &lt;&lt; 9); ++i) {\r\n            vector&lt;int&gt; cur;\r\n            int sum = 0;\r\n            \/\/ Use j if (j - 1)-th bit is 1\r\n            for (int j = 1; j &lt;= 9; ++j)\r\n                if (i &amp; (1 &lt;&lt; (j - 1))) {\r\n                    sum += j;\r\n                    cur.push_back(j);\r\n                }\r\n            if (sum == n &amp;&amp; cur.size() == k) \r\n                ans.push_back(cur);\r\n        }\r\n        \r\n        return ans;\r\n    }\r\n};<\/pre>\n<p>&nbsp;<\/p>\n<p>Python<\/p>\n<pre class=\"lang:c++ decode:true \">class Solution:\r\n    def combinationSum3(self, k, n):\r\n        def dfs(k, n, s, cur, ans):\r\n            if k == 0 and n == 0: \r\n                ans.append(list(cur))\r\n            for i in range(s, min(n + 1, 10)):\r\n                dfs(k - 1, n - i, i + 1, cur + [i], ans)\r\n        \r\n        ans = []        \r\n        dfs(k, n, 1, [], ans)\r\n        return ans\r\n<\/pre>\n<p>&nbsp;<\/p>\n<p>&nbsp;<\/p>\n<p><strong>Related problems:<\/strong><\/p>\n<ul>\n<li><a href=\"http:\/\/zxi.mytechroad.com\/blog\/searching\/leetcode-39-combination-sum\/\">[\u89e3\u9898\u62a5\u544a] LeetCode 39. Combination Sum<\/a><\/li>\n<li><a href=\"http:\/\/zxi.mytechroad.com\/blog\/searching\/leetcode-40-combination-sum-ii\/\">[\u89e3\u9898\u62a5\u544a] LeetCode 40. Combination Sum II<\/a><\/li>\n<\/ul>\n","protected":false},"excerpt":{"rendered":"<p>\u9898\u76ee\u5927\u610f\uff1a\u8f93\u51fa\u6240\u6709\u7528k\u4e2a\u6570\u7684\u548c\u4e3an\u7684\u7ec4\u5408\u3002\u53ef\u4ee5\u4f7f\u7528\u7684\u5143\u7d20\u662f1\u52309\u3002 Problem: Find all possible combinations of\u00a0k\u00a0numbers that add up to a number\u00a0n, given that only numbers from 1 to 9 can be used and&#8230;<\/p>\n","protected":false},"author":1,"featured_media":0,"comment_status":"open","ping_status":"open","sticky":false,"template":"","format":"standard","meta":{"footnotes":""},"categories":[164,44],"tags":[166,122,33,42,62],"class_list":["post-724","post","type-post","status-publish","format-standard","hentry","category-medium","category-searching","tag-backtracking","tag-combination","tag-dfs","tag-search","tag-sum","entry","simple"],"_links":{"self":[{"href":"https:\/\/zxi.mytechroad.com\/blog\/wp-json\/wp\/v2\/posts\/724","targetHints":{"allow":["GET"]}}],"collection":[{"href":"https:\/\/zxi.mytechroad.com\/blog\/wp-json\/wp\/v2\/posts"}],"about":[{"href":"https:\/\/zxi.mytechroad.com\/blog\/wp-json\/wp\/v2\/types\/post"}],"author":[{"embeddable":true,"href":"https:\/\/zxi.mytechroad.com\/blog\/wp-json\/wp\/v2\/users\/1"}],"replies":[{"embeddable":true,"href":"https:\/\/zxi.mytechroad.com\/blog\/wp-json\/wp\/v2\/comments?post=724"}],"version-history":[{"count":11,"href":"https:\/\/zxi.mytechroad.com\/blog\/wp-json\/wp\/v2\/posts\/724\/revisions"}],"predecessor-version":[{"id":2687,"href":"https:\/\/zxi.mytechroad.com\/blog\/wp-json\/wp\/v2\/posts\/724\/revisions\/2687"}],"wp:attachment":[{"href":"https:\/\/zxi.mytechroad.com\/blog\/wp-json\/wp\/v2\/media?parent=724"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/zxi.mytechroad.com\/blog\/wp-json\/wp\/v2\/categories?post=724"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/zxi.mytechroad.com\/blog\/wp-json\/wp\/v2\/tags?post=724"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}