{"id":4640,"date":"2019-01-14T20:50:26","date_gmt":"2019-01-15T04:50:26","guid":{"rendered":"https:\/\/zxi.mytechroad.com\/blog\/?p=4640"},"modified":"2019-01-14T20:53:00","modified_gmt":"2019-01-15T04:53:00","slug":"leetcode-976-largest-perimeter-triangle","status":"publish","type":"post","link":"https:\/\/zxi.mytechroad.com\/blog\/greedy\/leetcode-976-largest-perimeter-triangle\/","title":{"rendered":"\u82b1\u82b1\u9171 LeetCode 976. Largest Perimeter Triangle"},"content":{"rendered":"\n

Given an array A<\/code> of positive lengths, return the largest perimeter of a triangle with non-zero area<\/strong>, formed from 3 of these lengths.<\/p>\n\n\n\n

If it is impossible to form any triangle of non-zero area, return 0<\/code>.<\/p>\n\n\n\n

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

Input: <\/strong>[2,1,2]\nOutput: <\/strong>5\n<\/pre>\n\n\n\n
Example 2:<\/strong><\/p>\n\n\n\n
Input: <\/strong>[1,2,1]\nOutput: <\/strong>0\n<\/pre>\n\n\n\n
Example 3:<\/strong><\/p>\n\n\n\n
Input: <\/strong>[3,2,3,4]\nOutput: <\/strong>10\n<\/pre>\n\n\n\n
Example 4:<\/strong><\/p>\n\n\n\n
Input: <\/strong>[3,6,2,3]\nOutput: <\/strong>8\n<\/pre>\n\n\n\n
Note:<\/strong><\/p>\n\n\n\n
  1. 3 <= A.length <= 10000<\/code><\/li>
  2. 1 <= A[i] <= 10^6<\/code><\/li><\/ol>\n\n\n\n
    Solution: Greedy<\/strong><\/h2>\n\n\n\n
    Answer must be 3 consecutive numbers in the sorted array
    if A[i] >= A[i+1] + A[i+2], then A[i] >= A[i+j] + A[i+k],  1 < j < k
    if A[i] < A[i+1] + A[i+2], then A[i] + A[i+1] + A[i+2] is the answer<\/p>\n\n\n\n
    \n
    C++<\/h2>\n
    \n\n
    \nclass Solution {\npublic:\n  int largestPerimeter(vector& A) {\n    sort(rbegin(A), rend(A));\n    for (int i = 0; i < A.size() - 2; ++i)\n      if (A[i] < A[i + 1] + A[i + 2])\n        return A[i] + A[i + 1] + A[i + 2];\n    return 0;\n  }\n};\n<\/pre>\n<\/div><\/div>\n\n\n\n
     <\/p>\n","protected":false},"excerpt":{"rendered":"
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