Quesion: http://oj.leetcode.com/problems/4sum/

Question Name: 4Sum or 4-sum

The key point is to prune when further qualified combination is not possible.

1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27 28 29 30 31 32 33 34 35 36 37 38 39 40 41 42 43 44 45 46 47 48 49 50 51 52 53 54 55 56 57 58 59 60 | class Solution: # @return a list of lists of length 4, [[val1,val2,val3,val4]] def fourSum(self, num, target): result = [] num.sort() firstIndex = 0 while firstIndex < len(num)-3: secondIndex = firstIndex + 1 while secondIndex < len(num)-2: thirdIndex = secondIndex + 1 fourthIndex = len(num) - 1 while thirdIndex < fourthIndex: if num[firstIndex] + num[secondIndex] + num[thirdIndex] + num[thirdIndex] > target: # Already too larger, not possible to find any qualified # list afterwards break if num[firstIndex] + num[secondIndex] + num[fourthIndex] + num[fourthIndex] < target: # Already too small, not possible to find any qualified # list forwards break temp = num[firstIndex] + num[secondIndex] + num[thirdIndex] + num[fourthIndex] if temp == target: result.append([num[firstIndex], num[secondIndex], num[thirdIndex], num[fourthIndex]]) # Skip the duplicate numbers as num[thirdIndex] thirdIndex += 1 while thirdIndex < fourthIndex + 1 and num[thirdIndex] == num[thirdIndex-1]: thirdIndex += 1 # Skip the duplicate numbers as num[fourthIndex] fourthIndex -= 1 while fourthIndex > thirdIndex -1 and num[fourthIndex] == num[fourthIndex+1]: fourthIndex -= 1 elif temp > target: # Skip the duplicate numbers as num[fourthIndex] fourthIndex -= 1 while fourthIndex > thirdIndex -1 and num[fourthIndex] == num[fourthIndex+1]: fourthIndex -= 1 else: # Skip the duplicate numbers as num[thirdIndex] thirdIndex += 1 while thirdIndex < fourthIndex + 1 and num[thirdIndex] == num[thirdIndex-1]: thirdIndex += 1 # Skip the duplicate numbers as num[secondIndex] secondIndex += 1 while secondIndex < len(num)-1 and num[secondIndex] == num[secondIndex-1]: secondIndex += 1 # Skip the duplicate numbers as num[firstIndex] firstIndex += 1 while firstIndex < len(num) - 2 and num[firstIndex] == num[firstIndex-1]: firstIndex += 1 return result |

According to your code, your idea is the same like 3-sum problem?

Yes, it is.