算法练习Day6 （Leetcode/Python-哈希表）
454. 4Sum II
Given four integer arrays nums1, nums2, nums3, and nums4 all of length n, return the number of tuples (i, j, k, l) such that:
0 <= i, j, k, l < nnums1[i] + nums2[j] + nums3[k] + nums4[l] == 0
Example 1:
Input: nums1 = [1,2], nums2 = [-2,-1], nums3 = [-1,2], nums4 = [0,2] Output: 2 Explanation: The two tuples are: 1. (0, 0, 0, 1) -> nums1[0] + nums2[0] + nums3[0] + nums4[1] = 1 + (-2) + (-1) + 2 = 0 2. (1, 1, 0, 0) -> nums1[1] + nums2[1] + nums3[0] + nums4[0] = 2 + (-1) + (-1) + 0 = 0
思路：
比起昨天的2sum，就是多了一重循环，把每两组array的number之和的所有可能视为2sum里的一个array，再直接套用2sum的解法。
因为不用去重，所以更为简单。今天接下来的两道题都是需要去重的。
class Solution(object):
def fourSumCount(self, nums1, nums2, nums3, nums4):
"""
:type nums1: List[int]
:type nums2: List[int]
:type nums3: List[int]
:type nums4: List[int]
:rtype: int
"""
# It is easy since repetition is allowed.
sum12 = {}
count = 0
for i in nums1:
for j in nums2:
sum12[0-(i+j)] = sum12.get(0-(i+j),0) + 1
for i in nums3:
for j in nums4:
if j+i in sum12:
count += sum12[(i+j)]
return count
# computation complexity O(n^2)算法复杂度 O(n^2)
15. 3Sum
Given an integer array nums, return all the triplets [nums[i], nums[j], nums[k]] such that i != j, i != k, and j != k, and nums[i] + nums[j] + nums[k] == 0.
Notice that the solution set must not contain duplicate triplets.
Example 1:
Input: nums = [-1,0,1,2,-1,-4] Output: [[-1,-1,2],[-1,0,1]] Explanation: nums[0] + nums[1] + nums[2] = (-1) + 0 + 1 = 0. nums[1] + nums[2] + nums[4] = 0 + 1 + (-1) = 0. nums[0] + nums[3] + nums[4] = (-1) + 2 + (-1) = 0. The distinct triplets are [-1,0,1] and [-1,-1,2]. Notice that the order of the output and the order of the triplets does not matter.
思路：元素需要被输出，不止是计数，考虑用双指针。先对数组进行排序。从左到右选定第一个元素，然后以这个元素右侧隔壁的元素为left，排序数组的最右为right指针。接下来根据sum的大小调整左右指针就好。
注意去重，允许相同的元素出现在一个3sum tuple中，但不要出现一样的tuple。所以是在遍历左右指针的时候，跳过重复的。
注意提前结束的几个标志：if nums[i] > 0: return result #排序后第一个元素必须大于0.
while right > left # right的值一定大于等于left的，因为在之后的if elif else中right和left有可能相等，所以这里是right>left，而不是>=。
solution:
class Solution(object):
def threeSum(self, nums):
"""
:type nums: List[int]
:rtype: List[List[int]]
"""
nums = sorted(nums)
result = []
# i is the first(smallest) value of the 3 in the 3Sum
for i in range(len(nums)):
if nums[i] > 0:
return result
if i > 0 and nums[i] == nums[i-1]: # skip the same value to avoid repetation
continue
left = i+1
right = len(nums)-1
while right > left: # Once left right pointers change in the loop, this rule should be checked again to ensure the availability.
sum_ = nums[i] + nums[left] +nums[right]
if sum_ < 0:
left += 1
elif sum_ > 0:
right -= 1
else:
result.append([nums[i], nums[left], nums[right]])
# skip the repeated items
while right > left and nums[right] == nums[right-1]:
right -= 1
while right > left and nums[left] == nums[left+1]: # do not forget while right > left。
left += 1
right -= 1
left += 1
return result
字典法：
class Solution:
def threeSum(self, nums: List[int]) -> List[List[int]]:
result = []
nums.sort()
# 找出a + b + c = 0
# a = nums[i], b = nums[j], c = -(a + b)
for i in range(len(nums)):
# 排序之后如果第一个元素已经大于零，那么不可能凑成三元组
if nums[i] > 0:
break
if i > 0 and nums[i] == nums[i - 1]: #三元组元素a去重
continue
d = {}
for j in range(i + 1, len(nums)):
if j > i + 2 and nums[j] == nums[j-1] == nums[j-2]: # 三元组元素b去重
continue
c = 0 - (nums[i] + nums[j])
if c in d:
result.append([nums[i], nums[j], c])
d.pop(c) # 三元组元素c去重
else:
d[nums[j]] = j
return result算法复杂度均为O(n^2)
18. 4Sum
Given an array nums of n integers, return an array of all the unique quadruplets [nums[a], nums[b], nums[c], nums[d]] such that:
0 <= a, b, c, d < na,b,c, anddare distinct.nums[a] + nums[b] + nums[c] + nums[d] == target
You may return the answer in any order.
Example 1:
Input: nums = [1,0,-1,0,-2,2], target = 0 Output: [[-2,-1,1,2],[-2,0,0,2],[-1,0,0,1]]
思路：比起上一题3sum，就是多加一重循环。遍历头两个元素组合的所有可能性，然后再用双指针找接下来两个元素。
class Solution(object):
def fourSum(self, nums, target):
"""
:type nums: List[int]
:type target: int
:rtype: List[List[int]]
"""
# compare with 3sum，add a for-loop
nums.sort()
n = len(nums)
result = []
for i in range(n):
if nums[i] > target and nums[i] > 0 and target > 0: #early stop, not necessary
break
if i > 0 and nums[i] == nums[i-1]:
continue
for j in range(i+1, n):
if nums[i] + nums[j] > target and target > 0: #early stop, not necessary
break
if j > i+1 and nums[j] == nums[j-1]:
continue
left, right = j+1, n-1
while left < right:
s = nums[i] + nums[j] + nums[left] + nums[right]
if s == target:
result.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 s < target:
left += 1
else:
right -= 1
return result
383. Ransom Note
Given two strings ransomNote and magazine, return true if ransomNote can be constructed by using the letters from magazine and false otherwise.
Each letter in magazine can only be used once in ransomNote.
Example 1:
Input: ransomNote = "a", magazine = "b" Output: false
思路：这题和昨天的题目类似。不过magazine里letter可能在赎金信里并没有用到。
class Solution(object):
def canConstruct(self, ransomNote, magazine):
"""
:type ransomNote: str
:type magazine: str
:rtype: bool
"""
counts = {}
for c in magazine:
counts[c] = counts.get(c, 0) + 1
for c in ransomNote:
if c not in counts or counts[c] == 0:
return False
counts[c] -= 1
return True算法复杂度O(m+n)