{"id":5152,"date":"2019-05-06T22:22:16","date_gmt":"2019-05-07T05:22:16","guid":{"rendered":"https:\/\/zxi.mytechroad.com\/blog\/?p=5152"},"modified":"2019-05-06T22:40:45","modified_gmt":"2019-05-07T05:40:45","slug":"leetcode-838-push-dominoes","status":"publish","type":"post","link":"https:\/\/zxi.mytechroad.com\/blog\/searching\/leetcode-838-push-dominoes\/","title":{"rendered":"\u82b1\u82b1\u9171 LeetCode 838. Push Dominoes"},"content":{"rendered":"\n

here are N<\/code> dominoes in a line, and we place each domino vertically upright.<\/p>\n\n\n\n

In the beginning, we simultaneously push some of the dominoes either to the left or to the right.<\/p>\n\n\n\n

\"\"\/<\/figure>\n\n\n\n

After each second, each domino that is falling to the left pushes the adjacent domino on the left.<\/p>\n\n\n\n

Similarly, the dominoes falling to the right push their adjacent dominoes standing on the right.<\/p>\n\n\n\n

When a vertical domino has dominoes falling on it from both sides, it stays still due to the balance of the forces.<\/p>\n\n\n\n

For the purposes of this question, we will consider that a falling domino expends no additional force to a falling or already fallen domino.<\/p>\n\n\n\n

Given a string “S” representing the initial state. S[i] = 'L'<\/code>, if the i-th domino has been pushed to the left; S[i] = 'R'<\/code>, if the i-th domino has been pushed to the right; S[i] = '.'<\/code>, if the i<\/code>-th domino has not been pushed.<\/p>\n\n\n\n

Return a string representing the final state. <\/p>\n\n\n\n

Example 1:<\/strong><\/p>\n\n\n\n

Input: <\/strong>\".L.R...LR..L..\"\nOutput: <\/strong>\"LL.RR.LLRRLL..\"\n<\/pre>\n\n\n\n
Example 2:<\/strong><\/p>\n\n\n\n
Input: <\/strong>\"RR.L\"\nOutput: <\/strong>\"RR.L\"\nExplanation: <\/strong>The first domino expends no additional force on the second domino.\n<\/pre>\n\n\n\n
Note:<\/strong><\/p>\n\n\n\n
  1. 0 <= N <= 10^5<\/code><\/li>
  2. String dominoes<\/code> contains only 'L<\/code>‘, 'R'<\/code> and '.'<\/code><\/li><\/ol>\n\n\n\n
    Solution: Simulation<\/strong><\/p>\n\n\n\n
    Simulate the push process, record the steps from L and R for each domino.
    steps(L) ==  steps(R) => “.”
    steps(L) < steps(R) => “L”
    steps(L) > steps(R) => “R”<\/p>\n\n\n\n
    Time complexity: O(n)
    Space complexity: O(n)<\/p>\n\n\n\n
    \n
    C++<\/h2>\n
    \n\n
    \n\/\/ Author: Huahua, running time: 24 ms, 15.2 MB\nclass Solution {\npublic:\n  string pushDominoes(string D) {\n    const int n = static_cast(D.size());\n    vector L(n, INT_MAX), R(n, INT_MAX);    \n    \n    for (int i = 0; i < n; ++i)\n      if (D[i] == 'L') {\n        L[i] = 0;\n        for (int j = i - 1; j >= 0 && D[j] == '.'; --j)\n          L[j] = L[j + 1] + 1;\n      } else if (D[i] == 'R') {        \n        R[i] = 0;\n        for (int j = i + 1; j < n && D[j] == '.'; ++j)\n          R[j] = R[j - 1] + 1;\n      }\n    \n    for (int i = 0; i < n; ++i)\n      if (L[i] < R[i]) D[i] = 'L';\n      else if (L[i] > R[i]) D[i] = 'R';\n    \n    return D;\n  }\n};\n<\/pre>\n<\/div><\/div>\n","protected":false},"excerpt":{"rendered":"
    here are N dominoes in a line, and we place each domino vertically upright. In the beginning, we simultaneously push some of the dominoes either to the left…<\/p>\n","protected":false},"author":1,"featured_media":0,"comment_status":"open","ping_status":"open","sticky":false,"template":"","format":"standard","meta":{"footnotes":""},"categories":[44],"tags":[34,177,179],"class_list":["post-5152","post","type-post","status-publish","format-standard","hentry","category-searching","tag-bfs","tag-medium","tag-simulation","entry","simple"],"_links":{"self":[{"href":"https:\/\/zxi.mytechroad.com\/blog\/wp-json\/wp\/v2\/posts\/5152","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=5152"}],"version-history":[{"count":5,"href":"https:\/\/zxi.mytechroad.com\/blog\/wp-json\/wp\/v2\/posts\/5152\/revisions"}],"predecessor-version":[{"id":5158,"href":"https:\/\/zxi.mytechroad.com\/blog\/wp-json\/wp\/v2\/posts\/5152\/revisions\/5158"}],"wp:attachment":[{"href":"https:\/\/zxi.mytechroad.com\/blog\/wp-json\/wp\/v2\/media?parent=5152"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/zxi.mytechroad.com\/blog\/wp-json\/wp\/v2\/categories?post=5152"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/zxi.mytechroad.com\/blog\/wp-json\/wp\/v2\/tags?post=5152"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}