python实现leetcode之40. 组合总和 II

解题思路

先将candidates排序，数组很短，排序很快
然后看最小的元素candidates[0]
如果最小的元素大于等于target，就可以停止递归了
否则，组合包含两种情况
1.有第一项first，然后才是rest的组合
2.没有第一项，都是rest的组合

40. 组合总和 II

代码

cache = {}
class Solution:
    def combinationSum2(self, candidates: List[int], target: int) -> List[List[int]]:
        candidates.sort()
        return cb(candidates, target)


def cb(candidates, target):
    key = repr(candidates) + ':' + repr(target)
    if key in cache: return cache[key]
    if not candidates or candidates[0] > target:
        rtv = []
    elif candidates[0] == target:
        rtv = [[c] for c in candidates if c == target]
    else:  # candidates[0] < target
        first, *rest = candidates
        x = cb(rest, target)
        y = cb(rest, target-candidates[0])
        rtv = x + [[first, *item] for item in y]
    cache[key] = dedup(rtv)
    return cache[key]



def dedup(s):
    x = []
    for item in s:
        if item not in x:
            x.append(item)
    return x