{"id":796,"date":"2017-11-14T08:21:18","date_gmt":"2017-11-14T16:21:18","guid":{"rendered":"http:\/\/zxi.mytechroad.com\/blog\/?p=796"},"modified":"2018-09-04T08:16:57","modified_gmt":"2018-09-04T15:16:57","slug":"leetcode-321-create-maximum-number","status":"publish","type":"post","link":"https:\/\/zxi.mytechroad.com\/blog\/dynamic-programming\/leetcode-321-create-maximum-number\/","title":{"rendered":"\u82b1\u82b1\u9171 LeetCode 321. Create Maximum Number"},"content":{"rendered":"<p><iframe loading=\"lazy\" title=\"\u82b1\u82b1\u9171 LeetCode 321. Create Maximum Number - \u5237\u9898\u627e\u5de5\u4f5c EP107\" width=\"500\" height=\"375\" src=\"https:\/\/www.youtube.com\/embed\/YYduNJfzWaA?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><strong>Problem:<\/strong><\/p>\n<p>Given two arrays of length\u00a0<code>m<\/code>\u00a0and\u00a0<code>n<\/code>\u00a0with digits\u00a0<code>0-9<\/code>\u00a0representing two numbers. Create the maximum number of length\u00a0<code>k &lt;= m + n<\/code>\u00a0from digits of the two. The relative order of the digits from the same array must be preserved. Return an array of the\u00a0<code>k<\/code>\u00a0digits. You should try to optimize your time and space complexity.<\/p>\n<p><b>Example 1:<\/b><\/p>\n<p>nums1 =\u00a0<code>[3, 4, 6, 5]<\/code><br \/>\nnums2 =\u00a0<code>[9, 1, 2, 5, 8, 3]<\/code><br \/>\nk =\u00a0<code>5<\/code><br \/>\nreturn\u00a0<code>[9, 8, 6, 5, 3]<\/code><\/p>\n<p><b>Example 2:<\/b><\/p>\n<p>nums1 =\u00a0<code>[6, 7]<\/code><br \/>\nnums2 =\u00a0<code>[6, 0, 4]<\/code><br \/>\nk =\u00a0<code>5<\/code><br \/>\nreturn\u00a0<code>[6, 7, 6, 0, 4]<\/code><\/p>\n<p><b>Example 3:<\/b><\/p>\n<p>nums1 =\u00a0<code>[3, 9]<\/code><br \/>\nnums2 =\u00a0<code>[8, 9]<\/code><br \/>\nk =\u00a0<code>3<\/code><br \/>\nreturn\u00a0<code>[9, 8, 9]<\/code><\/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>\u9898\u76ee\u5927\u610f\uff1a\u7ed9\u4f60\u4e24\u4e2a\u6570\u5b57\u6570\u7ec4\u548ck\uff0c\u8fd4\u56de\u4ece\u4e24\u4e2a\u6570\u7ec4\u4e2d\u9009\u53d6k\u4e2a\u6570\u5b57\u80fd\u591f\u7ec4\u6210\u7684\u6700\u5927\u503c\u3002<\/p>\n<h1><strong>Idea:\u00a0<\/strong>Greedy + DP<\/h1>\n<p><a href=\"http:\/\/zxi.mytechroad.com\/blog\/wp-content\/uploads\/2017\/11\/321-ep107-2.png\"><img loading=\"lazy\" decoding=\"async\" class=\"alignnone wp-image-804 size-full\" src=\"http:\/\/zxi.mytechroad.com\/blog\/wp-content\/uploads\/2017\/11\/321-ep107-2.png\" alt=\"\" width=\"960\" height=\"540\" srcset=\"https:\/\/zxi.mytechroad.com\/blog\/wp-content\/uploads\/2017\/11\/321-ep107-2.png 960w, https:\/\/zxi.mytechroad.com\/blog\/wp-content\/uploads\/2017\/11\/321-ep107-2-300x169.png 300w, https:\/\/zxi.mytechroad.com\/blog\/wp-content\/uploads\/2017\/11\/321-ep107-2-768x432.png 768w, https:\/\/zxi.mytechroad.com\/blog\/wp-content\/uploads\/2017\/11\/321-ep107-2-624x351.png 624w\" sizes=\"auto, (max-width: 960px) 100vw, 960px\" \/><\/a><\/p>\n<h1><strong>Solution:<\/strong><\/h1>\n<p>Time complexity: O(k * (n1+n2)^2)<\/p>\n<p>Space complexity: O(n1+n2)<\/p>\n<div class=\"responsive-tabs\">\n<h2 class=\"tabtitle\">C++<\/h2>\n<div class=\"tabcontent\">\n\n<pre class=\"lang:c++ decode:true \">\/\/ Author: Huahua\r\n\/\/ Runtime: 26 ms\r\nclass Solution {\r\npublic:\r\n    vector&lt;int&gt; maxNumber(vector&lt;int&gt;&amp; nums1, vector&lt;int&gt;&amp; nums2, int k) {\r\n        vector&lt;int&gt; ans;\r\n        const int n1 = nums1.size();\r\n        const int n2 = nums2.size();\r\n        for (int i = max(0, k - n2); i &lt;= min(k, n1); ++i)\r\n            ans = max(ans, maxNumber(maxNumber(nums1, i), \r\n                                     maxNumber(nums2, k - i)));\r\n        return ans;\r\n    }\r\nprivate:    \r\n    vector&lt;int&gt; maxNumber(const vector&lt;int&gt;&amp; nums, int k) {\r\n        vector&lt;int&gt; ans(k);                \r\n        int j = 0;\r\n        for (int i = 0; i &lt; nums.size(); ++i) {\r\n            while (j &gt; 0 &amp;&amp; nums[i] &gt; ans[j - 1] \r\n                   &amp;&amp; nums.size() - i &gt; k - j) --j;\r\n            if (j &lt; k) ans[j++] = nums[i];\r\n        }\r\n        return ans;\r\n    }\r\n    \r\n    vector&lt;int&gt; maxNumber(const vector&lt;int&gt;&amp; nums1, const vector&lt;int&gt;&amp; nums2) {\r\n        vector&lt;int&gt; ans(nums1.size() + nums2.size());\r\n        auto s1 = nums1.cbegin();\r\n        auto e1 = nums1.cend();\r\n        auto s2 = nums2.cbegin();\r\n        auto e2 = nums2.cend();        \r\n        int index = 0;\r\n        while (s1 != e1 || s2 != e2)\r\n            ans[index++] = \r\n              lexicographical_compare(s1, e1, s2, e2) ? *s2++ : *s1++;\r\n        return ans;\r\n    }\r\n};<\/pre>\n\n<\/div><h2 class=\"tabtitle\">Java<\/h2>\n<div class=\"tabcontent\">\n\n<pre class=\"lang:java decode:true \">\/\/ Author: Huahua\r\n\/\/ Runtime: 18 ms\r\nclass Solution {\r\n    public int[] maxNumber(int[] nums1, int[] nums2, int k) {\r\n        int[] best = new int[0];\r\n        for (int i = Math.max(0, k - nums2.length); \r\n                 i &lt;= Math.min(k, nums1.length); ++i)            \r\n            best = max(best, 0, \r\n                       maxNumber(maxNumber(nums1, i), \r\n                                 maxNumber(nums2, k - i)), 0);\r\n        return best;\r\n    }\r\n    \r\n    private int[] maxNumber(int[] nums, int k) {\r\n        int[] ans = new int[k];\r\n        int j = 0;\r\n        for (int i = 0; i &lt; nums.length; ++i) {\r\n            while (j &gt; 0 &amp;&amp; nums[i] &gt; ans[j - 1] \r\n                &amp;&amp; nums.length - i &gt; k - j) --j;\r\n            if (j &lt; k)\r\n                ans[j++] = nums[i];\r\n        }        \r\n        return ans;\r\n    }\r\n    \r\n    private int[] maxNumber(int[] nums1, int[] nums2) {\r\n        int[] ans = new int[nums1.length + nums2.length];\r\n        int s1 = 0;\r\n        int s2 = 0;\r\n        int index = 0;\r\n        while (s1 != nums1.length || s2 != nums2.length)\r\n            ans[index++] = max(nums1, s1, nums2, s2) == nums1 ? \r\n                           nums1[s1++] : nums2[s2++];\r\n        return ans;\r\n    }\r\n    \r\n    private int[] max(int[] nums1, int s1, int[] nums2, int s2) {\r\n        for (int i = s1; i &lt; nums1.length; ++i) {\r\n            int j = s2 + i - s1;\r\n            if (j &gt;= nums2.length) return nums1;\r\n            if (nums1[i] &lt; nums2[j]) return nums2;\r\n            if (nums1[i] &gt; nums2[j]) return nums1;\r\n        }\r\n        return nums2;\r\n    }\r\n}<\/pre>\n<\/div><\/div>\n","protected":false},"excerpt":{"rendered":"<p>Problem: Given two arrays of length\u00a0m\u00a0and\u00a0n\u00a0with digits\u00a00-9\u00a0representing two numbers. Create the maximum number of length\u00a0k &lt;= m + n\u00a0from digits of the two. The relative&#8230;<\/p>\n","protected":false},"author":1,"featured_media":0,"comment_status":"open","ping_status":"open","sticky":false,"template":"","format":"standard","meta":{"footnotes":""},"categories":[46,51,165],"tags":[18,88,217,152],"class_list":["post-796","post","type-post","status-publish","format-standard","hentry","category-dynamic-programming","category-greedy","category-hard","tag-dp","tag-greedy","tag-hard","tag-max-number","entry","simple"],"_links":{"self":[{"href":"https:\/\/zxi.mytechroad.com\/blog\/wp-json\/wp\/v2\/posts\/796","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=796"}],"version-history":[{"count":12,"href":"https:\/\/zxi.mytechroad.com\/blog\/wp-json\/wp\/v2\/posts\/796\/revisions"}],"predecessor-version":[{"id":3860,"href":"https:\/\/zxi.mytechroad.com\/blog\/wp-json\/wp\/v2\/posts\/796\/revisions\/3860"}],"wp:attachment":[{"href":"https:\/\/zxi.mytechroad.com\/blog\/wp-json\/wp\/v2\/media?parent=796"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/zxi.mytechroad.com\/blog\/wp-json\/wp\/v2\/categories?post=796"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/zxi.mytechroad.com\/blog\/wp-json\/wp\/v2\/tags?post=796"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}