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Merge pull request #458 from EnzoSeason/leetcode-115
update115: 使用一维dp数组,并剪枝
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@ -186,6 +186,40 @@ class Solution:
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return dp[-1][-1]
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```
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Python3:
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```python
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class SolutionDP2:
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"""
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既然dp[i]只用到dp[i - 1]的状态,
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我们可以通过缓存dp[i - 1]的状态来对dp进行压缩,
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减少空间复杂度。
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(原理等同同于滚动数组)
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"""
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def numDistinct(self, s: str, t: str) -> int:
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n1, n2 = len(s), len(t)
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if n1 < n2:
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return 0
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dp = [0 for _ in range(n2 + 1)]
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dp[0] = 1
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for i in range(1, n1 + 1):
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# 必须深拷贝
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# 不然prev[i]和dp[i]是同一个地址的引用
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prev = dp.copy()
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# 剪枝,保证s的长度大于等于t
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# 因为对于任意i,i > n1, dp[i] = 0
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# 没必要跟新状态。
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end = i if i < n2 else n2
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for j in range(1, end + 1):
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if s[i - 1] == t[j - 1]:
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dp[j] = prev[j - 1] + prev[j]
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else:
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dp[j] = prev[j]
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return dp[-1]
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```
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Go:
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