An array of integers contains elements such that each element is 1 more or less than its preceding element. Now we are given a number, we need to determine the index of that number's first occurrence in the array. Need to optimize linear search. Its not homework.
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问题:
回答1:
My algorithm would be like this:
- p=0
- if (A[p] == x) then idx=p and the algorithm is finished else goto next step
- set p += |x-A[p]|
- goto 2.
say A[p] > x. Then, since A items increase or decrease by 1, idx is for sure at least (A[p] - x) index away from p. Same principle goes for A[p] < x.
int p=0, idx=-1; while (p<len && p>=0) if (A[p]==x) idx = p; else p+= abs(x-A[p]); Time complexity: The worst case would be O(n). I expect the average case to be better than O(n) ( I think it's O(log n) but not sure about it). Running time: Definitely better than linear search for all cases.
回答2:
You can't be faster than linear. The following code should be as fast as you can go:
int findFirstIndex(int num, int[] array) { int i = 0; while (i< array.length) if (array[i] == num) return i; else i += abs(array[i] - num) return -1 } But it's still O(array.length) in the worst case. Think for example if you are looking for 1 in an array containing only 0. Then, you can't skip any position.
回答3:
Start from first position; now consider the difference between the searched number N and first number. if array[0] == N then we finished; otherwise we have to jump of abs(array[0] -N ) positions; now just repeat this until the end of the array.