[LeetCode] 80. Remove Duplicates from Sorted Array II 有序數(shù)組中去除重復項之二

Given a sorted array nums, remove the duplicates in-place such that duplicates appeared at most twice and return the new length.

Do not allocate extra space for another array, you must do this by modifying the input array in-place with O(1) extra memory.

Example 1:

Given nums = [1,1,1,2,2,3],

Your function should return length =

5

, with the first five elements of

nums

being

1, 1, 2, 2

and 3 respectively.

It doesn't matter what you leave beyond the returned length.

Example 2:

Given nums = [0,0,1,1,1,1,2,3,3],

Your function should return length =

7

, with the first seven elements of

nums

being modified to 

0

, 0, 1, 1, 2, 3 and 3 respectively.

It doesn't matter what values are set beyond the returned length.

Clarification:

Confused why the returned value is an integer but your answer is an array?

Note that the input array is passed in by reference, which means modification to the input array will be known to the caller as well.

Internally you can think of this:

// nums is passed in by reference. (i.e., without making a copy)
int len = removeDuplicates(nums);

// any modification to nums in your function would be known by the caller.
// using the length returned by your function, it prints the first len elements.
for (int i = 0; i < len; i++) {
print(nums[i]);
}

這道題是之前那道 Remove Duplicates from Sorted Array 的拓展，這里允許最多重復的次數(shù)是兩次，那么可以用一個變量 cnt 來記錄還允許有幾次重復，cnt 初始化為1，如果出現(xiàn)過一次重復，則 cnt 遞減1，那么下次再出現(xiàn)重復，快指針直接前進一步，如果這時候不是重復的，則 cnt 恢復1，由于整個數(shù)組是有序的，所以一旦出現(xiàn)不重復的數(shù)，則一定比這個數(shù)大，此數(shù)之后不會再有重復項。理清了上面的思路，則代碼很好寫了：

解法一：

class Solution {
public:
    int removeDuplicates(vector<int>& nums) {
        int pre = 0, cur = 1, cnt = 1, n = nums.size();
        while (cur < n) {
            if (nums[pre] == nums[cur] && cnt == 0) ++cur;
            else {
                if (nums[pre] == nums[cur]) --cnt;
                else cnt = 1;
                nums[++pre] = nums[cur++];
            }
        }
        return nums.empty() ? 0 : pre + 1;
    }
};

這里其實也可以用類似于 Remove Duplicates from Sorted Array 中的解法三的模版，由于這里最多允許兩次重復，那么當前的數(shù)字 num 只要跟上上個覆蓋位置的數(shù)字 nusm[i-2] 比較，若 num 較大，則絕不會出現(xiàn)第三個重復數(shù)字（前提是數(shù)組是有序的），這樣的話根本不需要管 nums[i-1] 是否重復，只要將重復個數(shù)控制在2個以內(nèi)就可以了，參見代碼如下：

解法二：

class Solution {
public:
    int removeDuplicates(vector<int>& nums) {
        int i = 0;
        for (int num : nums) {
            if (i < 2 || num > nums[i - 2]) {
                nums[i++] = num;
            }
        }
        return i;
    }
};

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