Remove Duplicates from Sorted ListII
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Follow up for "Remove Duplicates":
What if duplicates are allowed at most twice?
For example,
Given sorted array A = [1,1,1,2,2,3]
,
Your function should return length = 5
, and A is now [1,1,2,2,3]
.
原始思路:
找一个标签数字,遍历向量元素,如果相等,则加1, 如果大于2,则continue遍历,如果不同 标签数字换成新的数字
#include<iostream>#include<vector>#include<algorithm>using namespace std;int main(){vector<int> a = { 1, 1, 1, 2, 2, 3, 3, 3, 4, 4, 4,4 };int k = 0;int key = a[0];int index = 0;int i = 0;while (i < a.size()){if (a[i] == key){k++;if (k<3) a[index++] = key;else {i++;continue;}}else{key = a[i];a[index++] = key;k = 1;}i++;}for (int i = 0; i < a.size(); i++)cout << a[i] << endl;system("pause");return 0;}
应为允许一个重复,所以判断就不是相邻而是相隔一个的数字是否不一样。
不一样就需要更新值,但注意到这题更新数组不能是当前的状态,而需要是更新前一个不同的数,用temp存当前值以便于下一次赋值。
#include<iostream>#include<vector>#include<algorithm>using namespace std;int main(){vector<int> a = { 1, 1, 1, 2, 2, 3, 3, 3, 4, 4, 4,4 };int k = 0;/*int key = a[0];int index = 0;int i = 0;while (i < a.size()){if (a[i] == key){k++;if (k<3) a[index++] = key;else {i++;continue;}}else{key = a[i];a[index++] = key;k = 1;}i++;}*/int num = 1, i, temp = a[1];for (i = 2; i<a.size(); ++i)if (a[i] != a[i - 2]){a[num++] = temp;temp = a[i];}a[num++] = temp;for (int i = 0; i < a.size(); i++)cout << a[i] << endl;system("pause");return 0;}
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