leetcode 330. 按要求补齐数组

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给定一个已排序的正整数数组 nums，和一个正整数 n 。从 [1, n] 区间内选取任意个数字补充到 nums 中，使得 [1, n] 区间内的任何数字都可以用 nums 中某几个数字的和来表示。请输出满足上述要求的最少需要补充的数字个数。

示例 1:

输入: nums = [1,3], n = 6
输出: 1 
解释:
根据 nums 里现有的组合 [1], [3], [1,3]，可以得出 1, 3, 4。
现在如果我们将 2 添加到 nums 中， 组合变为: [1], [2], [3], [1,3], [2,3], [1,2,3]。
其和可以表示数字 1, 2, 3, 4, 5, 6，能够覆盖 [1, 6] 区间里所有的数。
所以我们最少需要添加一个数字。

示例 2:

输入: nums = [1,5,10], n = 20
输出: 2
解释: 我们需要添加 [2, 4]。

示例 3:

输入: nums = [1,2,2], n = 5
输出: 0
  • mysolution 
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def greedy(nums,n):
    needed = set(range(1,n+1))#all need to 
    covered = set()#cover to all
    tmpcover = []
    full = []
    for i in nums:
        if i <= n:full.append(i)
    pick = needed-set(full)#can choose num
    for i in full:
        for j in covered:
            tmpcover.append(i+j)
        tmpcover.append(i)
        covered = set(tmpcover)
    needed -= covered
    count = 0
    while needed:
        state_covered = set()
        k = 0
        for i in pick:#find suitable picknum to maxsuit needed
            tmpcover = []
            for j in covered:
                tmpcover.append(i+j)
            tmpcover.append(i)
            state_covered = set(tmpcover)
            state_covered = needed&state_covered
            if len(state_covered) > k:
                k = len(state_covered)
                covered = state_covered
        needed -= covered
        count += 1
    return count
  • clear solution
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def greedy(nums,n):
    miss,added,i=1,0,0
    while miss <= n: 
        if i < len(nums) and nums[i] <= miss:
            miss += nums[i]
            i += 1
        else:
            miss += miss
            added += 1
    return added