@@ -254,12 +254,113 @@ class Solution:
254254 hashmap[num] = i
255255```
256256
257+ ### [ 3sum] ( https://leetcode.cn/problems/3sum/ )
258+ > 给你一个整数数组 nums ,判断是否存在三元组 [ nums[ i] , nums[ j] , nums[ k]] 满足 i != j、i != k 且 j != k ,同时还满足 nums[ i] + nums[ j] + nums[ k] == 0 。请
259+ > 你返回所有和为 0 且不重复的三元组。
260+
261+ > 注意:** 答案中不可以包含重复的三元组** 。
262+
263+ 思路:hash表的方法也能用,但使用双指针的解法更快
264+ 265+ ``` python
266+ class Solution :
267+ def threeSum (self nums : List[int ]) -> List[List[int ]]:
268+ if len (nums) < 3 : return []
269+ 270+ nums = sorted (nums)
271+ res = list ()
272+ i = 0
273+ while i < len (nums):
274+ if nums[i] > 0 : break
275+ target = 0 - nums[i]
276+ left, right = i + 1 , len (nums) - 1
277+ while left < right:
278+ sum2 = nums[left] + nums[right]
279+ if sum2 == target:
280+ res.append([nums[i], nums[left], nums[right]])
281+ while left < right and nums[left+ 1 ] == nums[left]:left += 1
282+ while right > left and nums[right- 1 ] == nums[right]: right -= 1
283+ left += 1
284+ right -= 1
285+ elif sum2 < target: left += 1
286+ else : right -= 1
287+ 288+ while i + 1 < len (nums) and nums[i+ 1 ] == nums[i]: i += 1
289+ i += 1
290+ 291+ return res
292+ ```
293+ 294+ ### [ 4sum] ( https://leetcode.cn/problems/4sum/ )
295+ > 给你一个由 n 个整数组成的数组 nums ,和一个目标值 target 。请你找出并返回满足下述全部条件且不重复的四元组 [ nums[ a] , nums[ b] , nums[ c] , nums[ d]] (若两个四元组元素一一>对应,则认为两个四元组重复):
296+
297+ > - 0 <= a, b, c, d < n
298+ > - a、b、c 和 d 互不相同
299+ > - nums[ a] + nums[ b] + nums[ c] + nums[ d] == target
300+ > 你可以按 任意顺序 返回答案 。
301+
302+ 跟上面的3sum一样,利用双指针进行转换,注意循环过程中剪枝和去重。
303+ 304+ ``` python
305+ class Solution :
306+ def fourSum (self nums : List[int ], target : int ) -> List[List[int ]]:
307+ if len (nums) < 4 : return []
308+ nums.sort()
309+ 310+ i = 0
311+ ans = list ()
312+ while i < len (nums) - 3 :
313+ if nums[i] * 4 > target: break
314+ tar3 = target - nums[i]
315+ j = i + 1
316+ while j < len (nums) - 2 :
317+ if nums[j] * 3 > tar3: break
318+ tar2 = tar3 - nums[j]
319+ left, right = j + 1 , len (nums)- 1
320+ while left < right:
321+ if nums[left] * 2 > tar2: break
322+ sum2 = nums[left] + nums[right]
323+ if sum2 == tar2:
324+ ans.append([nums[i], nums[j], nums[left], nums[right]])
325+ while left < right and nums[left+ 1 ] == nums[left]: left += 1
326+ while left < right and nums[right- 1 ] == nums[right]: right -= 1
327+ left += 1
328+ right -= 1
329+ elif sum2 < tar2: left += 1
330+ else : right -= 1
331+ while j + 1 < len (nums) and nums[j+ 1 ] == nums[j]: j+= 1
332+ j += 1
333+ 334+ while i + 1 < len (nums) and nums[i + 1 ] == nums[i]: i += 1
335+ i += 1
336+ 337+ return ans
338+ ```
339+ 257340### [ 4sum-ii] ( https://leetcode.cn/problems/4sum-ii/ )
258341> 给你四个整数数组 nums1、nums2、nums3 和 nums4 ,数组长度都是 n ,请你计算有多少个元组 (i, j, k, l) 能满足:
259342> - 0 <= i, j, k, l < n
260343> - nums1[ i] + nums2[ j] + nums3[ k] + nums4[ l] == 0
261344
345+ 思路:这个题目跟前面的3sum和4sum不一样,只让找出个数且没有去重逻辑,跟2sum比较类似,所以使用hash表比较方便
346+ 347+ ``` python
348+ from collections import defaultdict
349+ class Solution :
350+ def fourSumCount (self nums1 : List[int ], nums2 : List[int ], nums3 : List[int ], nums4 : List[int ]) -> int :
351+ res = 0
352+ n = len (nums1)
353+ hashmap = defaultdict(int )
354+ for i in range (n):
355+ for j in range (n):
356+ hashmap[nums1[i] + nums2[j]] += 1
357+ 358+ for i in range (n):
359+ for j in range (n):
360+ res += hashmap[- (nums3[i] + nums4[j])]
262361
362+ return res
363+ ```
263364
264365
265366### [ ransom-note] ( https://leetcode.cn/problems/ransom-note/ )
@@ -317,7 +418,7 @@ class Solution:
317418 if k not in m_d or m_d[k] < v: return False
318419
319420 return True
320- ```
421+ ```
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