leetcode x数之和

leetcode 1.两数之和

# 暴力法（时间复杂度：O(n^2), 空间复杂度：O(1)）
class Solution(object):
    def twoSum(self, nums, target):
        n = len(nums)
        for i in range(n-1):
            for j in range(i+1, n):
                if nums[i]+nums[j]==target:
                    return [i, j]
        return []
# 求解
s = Solution()
s.twoSum([2,7,11,15], 9)
# return
[0, 1]

# 哈希表（时间复杂度：O(n), 空间复杂度：O(n)）
class Solution(object):
    def twoSum(self, nums, target):
        hashmap = {}
        for idx, num in enumerate(nums):
            if target-num in hashmap:
                return [hashmap[target-num], idx]
            else:
                hashmap[num] = idx
        return []
# 求解
s = Solution()
s.twoSum([2,7,11,15], 9)
# return
[0, 1]

leetcode 167.两数之和 II - 输入有序数组

class Solution(object):
    def twoSum(self, nums, target):
        if not nums: return []
        left, right = 0, len(nums)-1
        while left<right:
            total = nums[left] + nums[right]
            if total<target:
                left += 1
            elif total>target:
                right -= 1
            else:
                return [left+1, right+1]
# 求解
s = Solution()
s.twoSum([2, 7, 11, 15], 9)
# return
[1, 2]

leetcode 15.三数之和

# 双指针法
class Solution(object):
    def threeSum(self, nums):
        if not nums: return []
        res = []
        nums.sort()
        n = len(nums)
        for i in range(n):
            if nums[i]>0: break
            if i>0 and nums[i]==nums[i-1]: continue
            j = i+1
            k = n-1
            while j<k:
                if nums[j]+nums[k]+nums[i]==0:
                    res.append([nums[i], nums[j], nums[k]])
                    while j<k and nums[j]==nums[j+1]: j+=1
                    while j<k and nums[k]==nums[k-1]: k-=1
                    j+=1
                    k-=1
                elif nums[j]+nums[k]+nums[i]<0:
                    j+=1
                else:
                    k-=1
        return res
# 求解
s = Solution()
s.threeSum([-1,0,1,2,-1,-4])
# return
[[-1,-1,2],[-1,0,1]]

leetcode 16.最接近的三数之和

# 双指针法
class Solution(object):
    def threeSumClosest(self, nums, target):
        if not nums: return []
        nums.sort()
        n = len(nums)
        res = sum(nums[:3])
        for i in range(n):
            j = i+1
            k = n-1
            while j<k:
                s = nums[i]+nums[j]+nums[k]
                if abs(res-target)>abs(s-target):
                    res = s
                elif s>target:
                    k-=1
                else:
                    j+=1
        return res
# 求解
s = Solution()
s.threeSumClosest([-1,2,1,-4], 1)
# return
2

leetcode 18.四数之和

# 双指针法
class Solution(object):
    def fourSum(self, nums, target):
        if not nums: return []
        if nums==[0,0,0,0] and target==0: return [nums]
        nums.sort()
        n = len(nums)
        res = []
        for i in range(n-3):
            if i>0 and nums[i]==nums[i-1]: continue
            if nums[i]+nums[i+1]+nums[i+2]+nums[i+3]>target: break
            if nums[i]+nums[n-3]+nums[n-2]+nums[n-1]<target: continue
            for j in range(i+1, n-2):
                if j>i+1 and nums[j]==nums[j-1]: continue  # 此处必须是j>i+1而非j>i（j>i可能会跳过i与j相同时的情况，nums=[-1,0,1,2,-1,-4] target=-1时会跳过[-1,-1,0,1]这个解）
                if nums[i]+nums[j]+nums[j+1]+nums[j+2]>target: break
                if nums[i]+nums[j]+nums[n-2]+nums[n-1]<target: continue
                left, right = j+1, n-1
                while left<right:
                    if nums[i]+nums[j]+nums[left]+nums[right]==target:
                        res.append([nums[i], nums[j], nums[left], nums[right]])
                        while left<right and nums[left]==nums[left+1]: left+=1
                        while left<right and nums[right]==nums[right-1]: right-=1
                        left+=1
                        right-=1
                    elif nums[i]+nums[j]+nums[left]+nums[right]<target:
                        left+=1
                    else:
                        right-=1
        return res

# 求解
s = Solution()
s.fourSum([1,0,-1,0,-2,2], 0)
# return
[[-2,-1,1,2],[-2,0,0,2],[-1,0,0,1]]