18. 4Sum; combination sum

18. 4Sum
class Solution(object):
    def fourSum(self, nums, target):
        nums.sort()
        results = []
        self.findNsum(nums, target, 4, [], results)
        return results

    def findNsum(self, nums, target, N, result, results):
        if len(nums) < N or N < 2: return

        # solve 2-sum
        if N == 2:
            l,r = 0,len(nums)-1
            while l < r:
                if nums[l] + nums[r] == target:
                    results.append(result + [nums[l], nums[r]])
                    l += 1
                    r -= 1
                    while l < r and nums[l] == nums[l - 1]:
                        l += 1
                    while r > l and nums[r] == nums[r + 1]:
                        r -= 1
                elif nums[l] + nums[r] < target:
                    l += 1
                else:
                    r -= 1
        else:
            for i in range(0, len(nums)-N+1):   # careful about range
                if target < nums[i]*N or target > nums[-1]*N:  # take advantages of sorted list
                    break
                if i == 0 or i > 0 and nums[i-1] != nums[i]:  # recursively reduce N
                    self.findNsum(nums[i+1:], target-nums[i], N-1, result+[nums[i]], results)

combination sum

def combinationSum(candidates, target):

    res = []

    candidates.sort()

    self.dfs(candidates, target, 0, [], res)

    return res

def dfs(self, nums, target, index, path, res):

    if target < 0:

        return  # backtracking

    if target == 0:

        res.append(path)

        return

    for i in xrange(index, len(nums)):

        self.dfs(nums, target-nums[i], i, path+[nums[i]], res)

    return result

转载于:https://www.cnblogs.com/ffeng0312/p/9739056.html