边工作边刷题：70天一遍leetcode: day 90-1

4 Sum

要点：

• recursion + k sum：可以用combination sum的思路，只要增加一个k的arg：因为k和i是联动的，所以recursion每层的k要-1
• 这题的success终止条件是k0 and target0，而failure的终止条件是k==0 or 过界 or target不符合条件。这里target不符合条件可以根据k来优化，类似Restore IP Addresses。利用nums是sorted，剩余的target的可能范围是[nums[i]N，nums[-1]N]，即最小/最大*N。如果target超过这个范围，就可以提前返回了。
• 优化：循环终止边界是len-k+1
• 这题如果只用recursion结束会TLE，而最后需要k=2然后用双指针。

class Solution(object):
    def fourSum(self, nums, target):
        """
        :type nums: List[int]
        :type target: int
        :rtype: List[List[int]]
        """
        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)
        return