80. Remove Duplicates from Sorted Array II

题目：

Follow up for "Remove Duplicates":
What if duplicates are allowed at most twice?

For example,
Given sorted array 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. It doesn't matter what you leave beyond the new length.

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Array Two Pointers

链接：  http://leetcode.com/problems/remove-duplicates-from-sorted-array-ii/

题解：

在排好序的数组里最多保留两个重复数字。设置一个limit，使用一个count，一个result用来计算最终结果。依照count和limit的关系决定是否移动到下一个index。

Time Complexity - O(n)， Space Complexity - O(1)。

public class Solution {
    public int removeDuplicates(int[] nums) {
        if(nums == null || nums.length == 0)
            return 0;
        int limit = 2, count = 1, result = 0;

        for(int i = 0; i < nums.length; i ++){
            if(i > 0 && nums[i] == nums[i - 1])
                count ++;
            else
                count = 1;
            if(count <= limit)
                nums[result ++] = nums[i];
        }
        return result;
    }
}

Update:

public class Solution {
    public int removeDuplicates(int[] nums) {
        if(nums == null || nums.length == 0)
            return 0;
        int count = 1, index = 1;

        for(int i = 1; i < nums.length; i++) {
            if(nums[i] == nums[i - 1])
                count ++;
            else
                count = 1;
            if(count <= 2)
                nums[index++] = nums[i];
        }

        return index;
    }
}

二刷：

就是设置一个limit，设置当前count为1，用来返回结果的index为1.

每次在循环里尝试更新count， 假如nums[i] = nums[i - 1]则count++，否则count = 1

在count <= limit的条件下，我们可以更新num[index++] = nums[i]。

最后返回index

Java:

Time Complexity - O(n)， Space Complexity - O(1)。

public class Solution {
    public int removeDuplicates(int[] nums) {
        int limit = 2, count = 1, index = 1;
        for (int i = 1; i < nums.length; i++) {
            count = nums[i] == nums[i - 1] ? count + 1 : 1;
            if (count <= limit) {
                nums[index++] = nums[i];
            }
        }
        return index;
    }
}

三刷:

几天没刷题，大脑反应速度就不够了，想得也细致，不能透过现象看本质。

Java:

public class Solution {
    public int removeDuplicates(int[] nums) {
        if (nums == null || nums.length == 0) return 0;
        int k = 2, count = 1, lo = 1;
        for (int i = 1; i < nums.length; i++) {
            if ((nums[i] == nums[i - 1] && count < k) || (nums[i] != nums[i - 1])) {
                count = (nums[i] == nums[i - 1]) ? count + 1 : 1 ;
                nums[lo++] = nums[i];
            }
        }
        return lo;
    }
}

Update:

使用二刷的逻辑

public class Solution {
    public int removeDuplicates(int[] nums) {
        if (nums == null || nums.length == 0) return 0;
        int limit = 2, count = 1, lo = 1;
        for (int i = 1; i < nums.length; i++) {
            count = (nums[i] == nums[i - 1]) ? count + 1 : 1 ;
            if (count <= limit) nums[lo++] = nums[i];
        }
        return lo;
    }
}

Update:

使用Stefan的代码

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

Reference:

https://leetcode.com/discuss/42348/3-6-easy-lines-c-java-python-ruby