{"id":339,"date":"2017-09-18T19:53:13","date_gmt":"2017-09-19T02:53:13","guid":{"rendered":"http:\/\/zxi.mytechroad.com\/blog\/?p=339"},"modified":"2018-07-10T18:41:21","modified_gmt":"2018-07-11T01:41:21","slug":"leetcode-221-maximal-square","status":"publish","type":"post","link":"https:\/\/zxi.mytechroad.com\/blog\/dynamic-programming\/leetcode-221-maximal-square\/","title":{"rendered":"\u82b1\u82b1\u9171 LeetCode 221. Maximal Square"},"content":{"rendered":"<p><iframe loading=\"lazy\" title=\"\u82b1\u82b1\u9171 LeetCode 221. Maximal Square - \u5237\u9898\u627e\u5de5\u4f5c EP62\" width=\"500\" height=\"375\" src=\"https:\/\/www.youtube.com\/embed\/vkFUB--OYy0?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 a 2D binary matrix filled with 0&#8217;s and 1&#8217;s, find the largest square containing only 1&#8217;s and return its area.<\/p>\n<p>For example, given the following matrix:<\/p>\n<p>1 0 1 0 0<br \/>\n1 0 <span style=\"color: red;\">1<\/span> <span style=\"color: red;\">1<\/span> 1<br \/>\n1 1 <span style=\"color: red;\">1<\/span> <span style=\"color: red;\">1<\/span> 1<br \/>\n1 0 0 1 0<\/p>\n<p>Return 4.<\/p>\n<p>&nbsp;<\/p>\n<p><strong>Idea:<\/strong><\/p>\n<p>Dynamic programming<\/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>Solution 0: O(n^5)<\/p>\n<p>&nbsp;<\/p>\n<p><a style=\"font-size: 1rem; color: #0f3647;\" href=\"http:\/\/zxi.mytechroad.com\/blog\/wp-content\/uploads\/2017\/09\/221-ep62-1.png\"><img loading=\"lazy\" decoding=\"async\" class=\"alignnone size-full wp-image-344\" src=\"http:\/\/zxi.mytechroad.com\/blog\/wp-content\/uploads\/2017\/09\/221-ep62-1.png\" alt=\"\" width=\"960\" height=\"540\" srcset=\"https:\/\/zxi.mytechroad.com\/blog\/wp-content\/uploads\/2017\/09\/221-ep62-1.png 960w, https:\/\/zxi.mytechroad.com\/blog\/wp-content\/uploads\/2017\/09\/221-ep62-1-300x169.png 300w, https:\/\/zxi.mytechroad.com\/blog\/wp-content\/uploads\/2017\/09\/221-ep62-1-768x432.png 768w, https:\/\/zxi.mytechroad.com\/blog\/wp-content\/uploads\/2017\/09\/221-ep62-1-624x351.png 624w\" sizes=\"auto, (max-width: 960px) 100vw, 960px\" \/><\/a><\/p>\n<p>Solution 1: O(n^3)<\/p>\n<p><a href=\"http:\/\/zxi.mytechroad.com\/blog\/wp-content\/uploads\/2017\/09\/221-ep62-2.png\"><img loading=\"lazy\" decoding=\"async\" class=\"alignnone size-full wp-image-343\" src=\"http:\/\/zxi.mytechroad.com\/blog\/wp-content\/uploads\/2017\/09\/221-ep62-2.png\" alt=\"\" width=\"960\" height=\"540\" srcset=\"https:\/\/zxi.mytechroad.com\/blog\/wp-content\/uploads\/2017\/09\/221-ep62-2.png 960w, https:\/\/zxi.mytechroad.com\/blog\/wp-content\/uploads\/2017\/09\/221-ep62-2-300x169.png 300w, https:\/\/zxi.mytechroad.com\/blog\/wp-content\/uploads\/2017\/09\/221-ep62-2-768x432.png 768w, https:\/\/zxi.mytechroad.com\/blog\/wp-content\/uploads\/2017\/09\/221-ep62-2-624x351.png 624w\" sizes=\"auto, (max-width: 960px) 100vw, 960px\" \/><\/a><\/p>\n<p><a style=\"font-size: 1rem;\" href=\"http:\/\/zxi.mytechroad.com\/blog\/wp-content\/uploads\/2017\/09\/221-ep62-3.png\"><img loading=\"lazy\" decoding=\"async\" class=\"alignnone size-full wp-image-342\" src=\"http:\/\/zxi.mytechroad.com\/blog\/wp-content\/uploads\/2017\/09\/221-ep62-3.png\" alt=\"\" width=\"960\" height=\"540\" srcset=\"https:\/\/zxi.mytechroad.com\/blog\/wp-content\/uploads\/2017\/09\/221-ep62-3.png 960w, https:\/\/zxi.mytechroad.com\/blog\/wp-content\/uploads\/2017\/09\/221-ep62-3-300x169.png 300w, https:\/\/zxi.mytechroad.com\/blog\/wp-content\/uploads\/2017\/09\/221-ep62-3-768x432.png 768w, https:\/\/zxi.mytechroad.com\/blog\/wp-content\/uploads\/2017\/09\/221-ep62-3-624x351.png 624w\" sizes=\"auto, (max-width: 960px) 100vw, 960px\" \/><\/a><\/p>\n<p>Solution 2: O(n^2)<\/p>\n<p><img loading=\"lazy\" decoding=\"async\" class=\"alignnone size-full wp-image-341\" style=\"font-size: 1rem;\" src=\"http:\/\/zxi.mytechroad.com\/blog\/wp-content\/uploads\/2017\/09\/221-ep62-4.png\" alt=\"\" width=\"960\" height=\"540\" srcset=\"https:\/\/zxi.mytechroad.com\/blog\/wp-content\/uploads\/2017\/09\/221-ep62-4.png 960w, https:\/\/zxi.mytechroad.com\/blog\/wp-content\/uploads\/2017\/09\/221-ep62-4-300x169.png 300w, https:\/\/zxi.mytechroad.com\/blog\/wp-content\/uploads\/2017\/09\/221-ep62-4-768x432.png 768w, https:\/\/zxi.mytechroad.com\/blog\/wp-content\/uploads\/2017\/09\/221-ep62-4-624x351.png 624w\" sizes=\"auto, (max-width: 960px) 100vw, 960px\" \/><\/p>\n<p>&nbsp;<\/p>\n<p><strong>Solution 1:<\/strong><\/p>\n<pre class=\"lang:c++ decode:true\">\/\/ Author: Huahua\r\n\/\/ Time complexity: O(n^3)\r\n\/\/ Running time: 39 ms\r\nclass Solution {\r\npublic:\r\n    int maximalSquare(vector&lt;vector&lt;char&gt;&gt;&amp; matrix) {\r\n        if (matrix.empty()) return 0;\r\n        int m = matrix.size();\r\n        int n = matrix[0].size();\r\n        \r\n        \/\/ sums[i][j] = sum(matrix[0][0] ~ matrix[i-1][j-1])\r\n        vector&lt;vector&lt;int&gt;&gt; sums(m + 1, vector&lt;int&gt;(n + 1, 0));\r\n        for (int i = 1; i &lt;= m; ++i)\r\n            for (int j = 1; j &lt;= n; ++j)        \r\n                sums[i][j] = matrix[i - 1][j - 1] - '0' \r\n                             + sums[i - 1][j]\r\n                             + sums[i][j - 1]\r\n                             - sums[i - 1][j - 1];\r\n        \r\n        int ans = 0;\r\n        for (int i = 1; i &lt;= m; ++i)\r\n            for (int j = 1; j &lt;= n; ++j)\r\n                for (int k = min(m - i + 1, n - j + 1); k &gt; 0; --k) {\r\n                    int sum = sums[i + k - 1][j + k - 1]\r\n                            - sums[i + k - 1][j - 1]\r\n                            - sums[i - 1][j + k - 1]\r\n                            + sums[i - 1][j - 1];\r\n                    \/\/ full of 1\r\n                    if (sum == k*k) {\r\n                        ans = max(ans, sum);\r\n                        break;\r\n                    }\r\n                }\r\n\r\n        return ans;\r\n    }\r\n};<\/pre>\n<p>&nbsp;<\/p>\n<p><strong>Solution 2:<\/strong><\/p>\n<pre class=\"lang:c++ decode:true  \">\/\/ Author: Huahua\r\n\/\/ Time complexity: O(n^2)\r\n\/\/ Running time: 6 ms\r\nclass Solution {\r\npublic:\r\n    int maximalSquare(vector&lt;vector&lt;char&gt;&gt;&amp; matrix) {\r\n        if (matrix.empty()) return 0;\r\n        int m = matrix.size();\r\n        int n = matrix[0].size();\r\n        \r\n        vector&lt;vector&lt;int&gt;&gt; sizes(m, vector&lt;int&gt;(n, 0));\r\n        \r\n        int ans = 0;\r\n        \r\n        for (int i = 0; i &lt; m; ++i)\r\n            for (int j = 0; j &lt; n; ++j) {\r\n                sizes[i][j] = matrix[i][j] - '0';\r\n                if (!sizes[i][j]) continue;                            \r\n                \r\n                if (i == 0 || j == 0) {\r\n                    \/\/ do nothing\r\n                } else if (i == 0)\r\n                    sizes[i][j] = sizes[i][j - 1] + 1;\r\n                else if (j == 0)\r\n                    sizes[i][j] = sizes[i - 1][j] + 1;\r\n                else\r\n                    sizes[i][j] = min(min(sizes[i - 1][j - 1], \r\n                                          sizes[i - 1][j]),\r\n                                          sizes[i][j - 1]) + 1;\r\n                \r\n                ans = max(ans, sizes[i][j]*sizes[i][j]);\r\n            }\r\n        return ans;\r\n    }\r\n};<\/pre>\n<p>&nbsp;<\/p>\n","protected":false},"excerpt":{"rendered":"<p>Problem: Given a 2D binary matrix filled with 0&#8217;s and 1&#8217;s, find the largest square containing only 1&#8217;s and return its area. For example, given&#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],"tags":[18,216,141,62],"class_list":["post-339","post","type-post","status-publish","format-standard","hentry","category-dynamic-programming","tag-dp","tag-matrix","tag-max","tag-sum","entry","simple"],"_links":{"self":[{"href":"https:\/\/zxi.mytechroad.com\/blog\/wp-json\/wp\/v2\/posts\/339","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=339"}],"version-history":[{"count":8,"href":"https:\/\/zxi.mytechroad.com\/blog\/wp-json\/wp\/v2\/posts\/339\/revisions"}],"predecessor-version":[{"id":3068,"href":"https:\/\/zxi.mytechroad.com\/blog\/wp-json\/wp\/v2\/posts\/339\/revisions\/3068"}],"wp:attachment":[{"href":"https:\/\/zxi.mytechroad.com\/blog\/wp-json\/wp\/v2\/media?parent=339"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/zxi.mytechroad.com\/blog\/wp-json\/wp\/v2\/categories?post=339"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/zxi.mytechroad.com\/blog\/wp-json\/wp\/v2\/tags?post=339"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}