第六周:[leetcode] 多思路解 169.Majority Element
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题目链接:链接
问题描述:
找到存在n个元素的数组中出现次数大于ceil(n//2)的主元素,即Majority Element;
该题难度为easy,比较简单,但通过参考学习,发现解题思路多且有趣,能开拓思维,因此记录此文(5种方法)。
solution1:
hash建表遍历数组(),记录元素出现次数,当次数count > ceil(n//2),得到答案;
int majorityElement(vector<int>& nums) { unordered_map<int, int> counts; int n = nums.size(); for (int i = 0; i < n; i++) if (++counts[nums[i]] > n / 2) return nums[i]; }
solution2:
对数组所有元素进行排序,则数组中第ceil(n//2)个数必为主元素;
int majorityElement(vector<int>& nums) { sort(nums.begin(), nums.end()); return nums[nums.size() / 2]; }
solution3:
️投票算法(Moore Voting Algorithm),利用主元素数量大于ceil(n//2)的条件进行投票,时间复杂度为O(n),对该类问题具有通用性:ceil(n/m);
int majorityElement(vector<int>& nums) { int major, counts = 0, n = nums.size(); for (int i = 0; i < n; i++) { if (!counts) { major = nums[i]; counts = 1; } //投票 else counts += (nums[i] == major) ? 1 : -1; } return major; }
solution4:
随机化算法,因为major ele出现的频率很高,每次选中机率大于1/2,也比较快得出答案;
int majorityElement(vector<int>& nums) { int n = nums.size(); srand(unsigned(time(NULL)));//随机数的生成 while (true) { int idx = rand() % n; int candidate = nums[idx]; int counts = 0; for (int i = 0; i < n; i++) if (nums[i] == candidate) counts++; if (counts > n / 2) return candidate; } }
solution5:
️位运算,出现次数大于n/2的数字(int 类型8个字节,32位),其每一位出现次数,必大于n/2,针对出现次数大于n/2的数字进行位复制,得出答案。
int majorityElement(vector<int>& nums) { int major = 0, n = nums.size(); for (int i = 0, mask = 1; i < 32; i++, mask <<= 1) { int bitCounts = 0; for (int j = 0; j < n; j++) { if (nums[j] & mask) bitCounts++; if (bitCounts > n / 2) { major |= mask; break; } } } return major; }
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