【leetcode题解】【再看一遍】【86】【M】Contains Duplicate III

Given an array of integers, find out whether there are two distinct indices i and j in the array such that the difference between nums[i] and nums[j] is at most t and the difference between i and j is at most k.

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The idea is like the bucket sort algorithm. Suppose we have consecutive buckets covering the range of nums with each bucket a width of (t+1). If there are two item with difference <= t, one of the two will happen:

(1) the two in the same bucket(2) the two in neighbor buckets

Note that we do not need to actually allocate a lot of buckets. At any time there will only be at most min(k, n) buckets. All we need to do is calculate the label of the bucket m = value/(t+1), and check the buckets m - 1, m, m + 1. The whole algorithm is then O(n).

借鉴桶排序的的思想+滑动窗口

如果： | nums[i] - nums[j] | <= t   式a

等价： | nums[i] / t - nums[j] / t | <= 1   式b

推出： | floor(nums[i] / t) - floor(nums[j] / t) | <= 1   式c

​等价： floor(nums[j] / t) ∈ {floor(nums[i] / t) - 1, floor(nums[i] / t), floor(nums[i] / t) + 1} 式d

运行结果长这个样子

Your input

[1,3,111,11114,111115,1111111,11111112,11111113]31

Your stdout

{}
{0: 1}
{0: 1, 1: 3}
{0: 1, 1: 3, 55: 111}
{0: 1, 1: 3, 5557: 11114, 55: 111}
{1: 3, 55557: 111115, 5557: 11114, 55: 111}
{555555: 1111111, 55557: 111115, 5557: 11114, 55: 111}

{555555: 1111111, 5555556: 11111112, 55557: 111115, 5557: 11114}

注释了的是原来的算法，想法不对，又太笨。

class Solution(object):    def containsNearbyAlmostDuplicate(self, nums, k, t):        if t < 0:            return False        l = len(nums)        d = {}        t += 1        # 1/3 = 0        #-1/3 = -1        for i in xrange(l):            print d            if i > k:                del d[nums[i - k - 1] / t]# 维护一个大小为k+1的桶            m = nums[i] / t            if m in d:                return True            elif m-1 in d and abs(nums[i] - d[m-1]) < t:                return True            elif m+1 in d and abs(nums[i] - d[m+1]) < t:                return True            d[m] = nums[i]        return False        '''        l = len(nums)        if not nums:            return 0 < k        if l == 1 or k == 0 or t < 0:            return False        if l <= k:            for i in range(0,l-1):                if abs(nums[i+1] - nums[i]) <= t:                    return True            return False        i = k - 1        df = [0]*(k )        temp = nums[:k]        temp.sort()        for ii in range(0,k):            print ii            if ii == l-1:                break            df[ii] = (temp[ii+1]-temp[ii])        #df.sort()        print df        minn = min(df)        if minn <= t:            return True        print df        while i < l-1:            i += 1            temp_df = abs(nums[i]-nums[i-1])            if temp_df < minn:                df.remove(minn)                minn = temp_df                df.append(minn)            #print df            if minn <= t:                return True        return False        '''