Saturday, November 15, 2014

[LeetCode] Remove Duplicates from Sorted Array I, II

Remove Duplicates from Sorted Array I
Given a sorted array, remove the duplicates in place such that each element appear only once and return the new length.
Do not allocate extra space for another array, you must do this in place with constant memory.
For example,
Given input array A = [1,1,2],
Your function should return length = 2, and A is now [1,2].

Remove Duplicates from Sorted Array II
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].


思路:Remove Duplicates from Sorted Array I
双指针,一个指针 i 扫描数组,一个指针 end 记录没有重复数字的新数组的尾部。
(1) A[end]=A[i],A[i]为重复数字,跳过。
(2) A[end]!=A[i],将A[i]放到A[end+1]位置,end++

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class Solution {
public:
    int removeDuplicates(int A[], int n) {
        if(n<2) return n;
        int end=0;
        for(int i=1; i<n; i++) {
            if(A[i]!=A[end]) {
                end++;
                if(i!=end) A[end] = A[i];
            }
        }
        return end+1;
    }
};


思路:Remove Duplicates from Sorted Array II

和I一样的思路,区别仅仅在于当A[end-1] = A[end] = A[i]时,A[i]为重复元素需跳过。而实际只需要比较A[end-1]和A[i],因为当A[end-1] = A[i]时,根据sorted arry特性必然也有A[end]=A[end-1]。


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class Solution {
public:
    int removeDuplicates(int A[], int n) {
        if(n<3) return n;
        int end = 1;
        for(int i=2; i<n; i++) {
            if(A[i]!=A[end-1]) 
                A[++end] = A[i];
        }
        return end+1;
    }
};

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