egin(); while (it != hash.end()) { if (--(it->second) == 0) { hash.erase(it++); } else { ++it; } } } } // Resets hash for the following counting. for (auto& it : hash) { it.second = 0; } // Counts the occurrence of each candidate integer. for (const auto& i : nums) { auto it = hash.find(i); if (it != hash.end()) { ++it->second; } } // Selects the integer which occurs > n / k times. vector ret; for (const pair& it : hash) { if (it.second > static_cast(n) / k) { ret.emplace_back(it.first); } } return ret[0]; } }; 4:["$","div",null,{"children":[["$","header",null,{"children":[["$","div",null,{"className":"text-3xl text-bold p-0 mt-2 mb-2","children":["$","h1",null,{"children":["Majority Number III",["$","span",null,{"className":"font-mono text-sm text-gray-500","children":"(LintCode 48):"}]]}]}],""]}],["$","article",null,{"children":[["$","div",null,{"className":"p-0","dangerouslySetInnerHTML":{"__html":"

Problem description: Visit and Support LintCode Search via
Difficulty: Medium
Tags: LintCode, Array, O(n), O(k), Medium,

Problem