leetcode 80. Remove Duplicates from Sorted Array II
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/*leetcode 80. Remove Duplicates from Sorted Array II题目大意:从排序好的数组中删除重复元素,最多允许重复次数为2次解题思路:对于排序的数组,如果重复次数大于2,那么一定是nums[i] == nums[i-2],即与前两个相等譬如说:1,1,1,2,2,2,3,4,5。(序号从0开始)第2个和第0个相等,故而说明是重复元素。在对数组扫描的过程中,要记录数组中删除多余元素后的序号,并将这个序号作为数组的序号,其他删除的元素直接覆盖就行了。*/#include <iostream>#include <string>#include <vector>using namespace std;class Solution {public: int removeDuplicates(vector<int>& nums) { int ret = 2; if (nums.size() < 2) return nums.size(); for (size_t i = 2; i < nums.size(); ++i) { if (nums[i] != nums[ret - 2]) //判断重复的次数是不是大于2 nums[ret++] = nums[i]; //直接覆盖,用ret记录新的位置 } return ret; }};void test_removeDuplicates(){ Solution sol; vector<int> nums{ 1,1,1,2,2,2,2,2,3,3,4,5,6,6,6,7,7,8,9,10 }; int ret = sol.removeDuplicates(nums); cout << ret << endl; for (int i = 0; i < ret; ++i) cout << nums[i] << " "; cout << endl;}int main(){ test_removeDuplicates(); return 0;}
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