{"id":1665,"date":"2018-01-25T22:08:15","date_gmt":"2018-01-26T06:08:15","guid":{"rendered":"http:\/\/zxi.mytechroad.com\/blog\/?p=1665"},"modified":"2018-04-21T18:04:34","modified_gmt":"2018-04-22T01:04:34","slug":"leetcode-763-partition-labels","status":"publish","type":"post","link":"https:\/\/zxi.mytechroad.com\/blog\/string\/leetcode-763-partition-labels\/","title":{"rendered":"\u82b1\u82b1\u9171 LeetCode 763. Partition Labels"},"content":{"rendered":"<p><iframe loading=\"lazy\" title=\"\u82b1\u82b1\u9171 LeetCode 763. Partition Labels - \u5237\u9898\u627e\u5de5\u4f5c EP161\" width=\"500\" height=\"375\" src=\"https:\/\/www.youtube.com\/embed\/s-1W5FDJ0lw?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\u628a\u5b57\u7b26\u4e32\u5206\u5272\u6210\u5c3d\u91cf\u591a\u7684\u4e0d\u91cd\u53e0\u5b50\u4e32\uff0c\u8f93\u51fa\u5b50\u4e32\u7684\u957f\u5ea6\u6570\u7ec4\u3002\u8981\u6c42\u76f8\u540c\u5b57\u7b26\u53ea\u80fd\u51fa\u73b0\u5728\u4e00\u4e2a\u5b50\u4e32\u4e2d\u3002<\/p>\n<p><strong>Problem:<\/strong><\/p>\n<p>A string\u00a0<code>S<\/code>\u00a0of lowercase letters is given. We want to partition this string into as many parts as possible so that each letter appears in at most one part, and return a list of integers representing the size of these parts.<\/p>\n<p><b>Example 1:<\/b><\/p>\n<pre class=\"\">Input: S = \"ababcbacadefegdehijhklij\"\r\nOutput: [9,7,8]\r\nExplanation:\r\nThe partition is \"ababcbaca\", \"defegde\", \"hijhklij\".\r\nThis is a partition so that each letter appears in at most one part.\r\nA partition like \"ababcbacadefegde\", \"hijhklij\" is incorrect, because it splits S into less parts.<\/pre>\n<p><img loading=\"lazy\" decoding=\"async\" class=\"alignnone size-full wp-image-1680\" src=\"http:\/\/zxi.mytechroad.com\/blog\/wp-content\/uploads\/2018\/01\/763-ep161.png\" alt=\"\" width=\"960\" height=\"540\" srcset=\"https:\/\/zxi.mytechroad.com\/blog\/wp-content\/uploads\/2018\/01\/763-ep161.png 960w, https:\/\/zxi.mytechroad.com\/blog\/wp-content\/uploads\/2018\/01\/763-ep161-300x169.png 300w, https:\/\/zxi.mytechroad.com\/blog\/wp-content\/uploads\/2018\/01\/763-ep161-768x432.png 768w\" sizes=\"auto, (max-width: 960px) 100vw, 960px\" \/><\/p>\n<p><strong>Solution 0: Brute Force<\/strong><\/p>\n<p>Time complexity: O(n^2)<\/p>\n<p>Space complexity: O(1)<\/p>\n<p>C++<\/p>\n<pre class=\"lang:default decode:true \">\/\/ Author: Huahua\r\n\/\/ Running time: 15 ms\r\nclass Solution {\r\npublic:\r\n    vector&lt;int&gt; partitionLabels(string S) {\r\n      vector&lt;int&gt; ans;\r\n      size_t start = 0;\r\n      size_t end = 0;\r\n      for (size_t i = 0; i &lt; S.size(); ++i) {\r\n        end = max(end, S.find_last_of(S[i]));\r\n        if (i == end) {\r\n          ans.push_back(end - start + 1);\r\n          start = end + 1;\r\n        }\r\n      }\r\n      return ans;\r\n    }\r\n};<\/pre>\n<p>Python<\/p>\n<pre class=\"lang:python decode:true  \">\"\"\"\r\nAuthor: Huahua\r\nRunning time : 48 ms\r\n\"\"\"\r\nclass Solution:\r\n  def partitionLabels(self, S):\r\n    ans = []\r\n    start, end = 0, 0\r\n    for i in range(len(S)):\r\n      end = max(end, S.rfind(S[i]))\r\n      if i == end:\r\n        ans.append(end - start + 1)\r\n        start = end + 1\r\n    return ans\r\n<\/pre>\n<p>&nbsp;<\/p>\n<p><strong>Solution 1: Greedy <\/strong><\/p>\n<p>Time complexity: O(n)<\/p>\n<p>Space complexity: O(26\/128)<\/p>\n<p>C++<\/p>\n<pre class=\"lang:c++ decode:true\">\/\/ Author: Huahua\r\n\/\/ Running time: 6 ms\r\nclass Solution {\r\npublic:\r\n    vector&lt;int&gt; partitionLabels(string S) {\r\n      vector&lt;int&gt; last_index(128, 0);\r\n      for (int i = 0; i &lt; S.size(); ++i)\r\n        last_index[S[i]] = i;\r\n      vector&lt;int&gt; ans;\r\n      int start = 0;\r\n      int end = 0;\r\n      for (int i = 0; i &lt; S.size(); ++i) {\r\n        end = max(end, last_index[S[i]]);\r\n        if (i == end) {\r\n          ans.push_back(end - start + 1);\r\n          start = end + 1;\r\n        }\r\n      }\r\n      return ans;\r\n    }\r\n};<\/pre>\n<p>Java<\/p>\n<pre class=\"lang:java decode:true \">\/\/ Author: Huahua\r\n\/\/ Running time: 16 ms\r\nclass Solution {\r\n  public List&lt;Integer&gt; partitionLabels(String S) {\r\n    int lastIndex[] = new int[128];\r\n    for (int i = 0; i &lt; S.length(); ++i)\r\n      lastIndex[(int)S.charAt(i)] = i;\r\n    List&lt;Integer&gt; ans = new ArrayList&lt;&gt;();\r\n    int start = 0;\r\n    int end = 0;\r\n    for (int i = 0; i &lt; S.length(); ++i) {\r\n      end = Math.max(end, lastIndex[(int)S.charAt(i)]);\r\n      if (i == end) {\r\n        ans.add(end - start + 1);\r\n        start = end + 1;\r\n      }\r\n    }\r\n    return ans;\r\n  }\r\n}<\/pre>\n<p>Python3<\/p>\n<pre class=\"lang:python decode:true \">\"\"\"\r\nAuthor: Huahua\r\nRunning time: 78 ms\r\n\"\"\"\r\nclass Solution:\r\n  def partitionLabels(self, S):\r\n    last_index = {}\r\n    for i, ch in enumerate(S):\r\n      last_index[ch] = i\r\n    start = end = 0\r\n    ans = []\r\n    for i, ch in enumerate(S):\r\n      end = max(end, last_index[ch])\r\n      if i == end:\r\n        ans.append(end - start + 1)\r\n        start = end + 1\r\n    return ans<\/pre>\n<p>&nbsp;<\/p>\n","protected":false},"excerpt":{"rendered":"<p>\u9898\u76ee\u5927\u610f\uff1a\u628a\u5b57\u7b26\u4e32\u5206\u5272\u6210\u5c3d\u91cf\u591a\u7684\u4e0d\u91cd\u53e0\u5b50\u4e32\uff0c\u8f93\u51fa\u5b50\u4e32\u7684\u957f\u5ea6\u6570\u7ec4\u3002\u8981\u6c42\u76f8\u540c\u5b57\u7b26\u53ea\u80fd\u51fa\u73b0\u5728\u4e00\u4e2a\u5b50\u4e32\u4e2d\u3002 Problem: A string\u00a0S\u00a0of lowercase letters is given. We want to partition this string into as many parts as possible so that each letter appears&#8230;<\/p>\n","protected":false},"author":1,"featured_media":0,"comment_status":"open","ping_status":"open","sticky":false,"template":"","format":"standard","meta":{"footnotes":""},"categories":[51,164,47],"tags":[88,150,4],"class_list":["post-1665","post","type-post","status-publish","format-standard","hentry","category-greedy","category-medium","category-string","tag-greedy","tag-partition","tag-string","entry","simple"],"_links":{"self":[{"href":"https:\/\/zxi.mytechroad.com\/blog\/wp-json\/wp\/v2\/posts\/1665","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=1665"}],"version-history":[{"count":8,"href":"https:\/\/zxi.mytechroad.com\/blog\/wp-json\/wp\/v2\/posts\/1665\/revisions"}],"predecessor-version":[{"id":2752,"href":"https:\/\/zxi.mytechroad.com\/blog\/wp-json\/wp\/v2\/posts\/1665\/revisions\/2752"}],"wp:attachment":[{"href":"https:\/\/zxi.mytechroad.com\/blog\/wp-json\/wp\/v2\/media?parent=1665"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/zxi.mytechroad.com\/blog\/wp-json\/wp\/v2\/categories?post=1665"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/zxi.mytechroad.com\/blog\/wp-json\/wp\/v2\/tags?post=1665"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}