Leetcode 15 three sum two pointer

使用 set 作為ans_set 可避免重複數組

enumerate python 技巧

list1[1,3,5,7,9]
for index, num in enumerate(list1):
    print('index:',i , 'num:',num)

My Solution 時間O(n^2)

def threeSum(self, nums):
    # 先將輸入數列排序後呈 [-4,-1,-1,0,1,2]
    nums = sorted(nums)
    length = len(nums)
    sol_list = set()
    if length < 3:
        return sol_list
    for index in range(length): 
        pivot_index = index
        if nums[pivot_index] > 0: # 不可能相加 = 0
            return sol_list
        if index >=1 and nums[index-1] == nums[index]: 
            continue
        # 設定其餘數列相加pivot 為 0
        target = 0 - nums[index]
        target_map = {}  # 利用dict 去尋找有無適合值 
        for pivot_index in range(pivot_index+1,length):
            if target_map.__contains__(target-nums[pivot_index]):
                #如果dict內有key 符合 target-nums[pivot_index]
                #代表條件符合新增於ans_list
                sol_list.add((nums[index],
                target-nums[pivot_index],nums[pivot_index]))
            else:  # 紀錄值在target_map內利後續查詢 
                target_map[nums[pivot_index]] = pivot_index 
    return sol_list

Best Solution Two pointer
時間O(n^2)

def threeSum(self, nums):
    # 先將輸入數列排序後呈 [-4,-1,-1,0,1,2]
    nums = sorted(nums)
    length = len(nums)
    sol_list = set()
    if length < 3:
        return sol_list
    for index in range(length-2):
        # 因為是已經排序過，所以nums[pivot_index] > 0 
        # 就不用繼續
        if nums[index] > 0:
            break
        if index >=1 and nums[index-1] == nums[index]: 
            continue
        # 想法為有兩個指標 l , r 
        # 當 sum 過大 r-1,sum過小 l+1 
        l = index + 1 # 左邊index
        r = length - 1 # 右邊index
        while l < r: # 以 index 為主key 去嘗試 l & r 的值
            sum = nums[index] + nums[l]+ nums[r]
            if sum < 0 : # 代表 left的值負太多 往右移
                l += 1
            elif sum > 0: # 代表 right 的值正太多 往左移
                r -= 1
            else:
                sol_list.add((nums[index],nums[l],nums[r])) 
                l+=1; r-=1
    return sol_list

tags: LeetCode three sum two pointer