leetcode sum相关算法题

1. Two Sum(https://oj.leetcode.com/problems/two-sum/)

解题思路：

解法一： 暴力，O(n2)时间复杂度，TLE

解法二：利用hash, 记录下数组中每个值对应的下标，再遍历一遍数组，通过查看target-num[i]的值是否在map中来确定另一个数值。时间复杂度O(n)

解法三：对num数组排序，O(nlog(n)), 然后左右夹逼O(n). 但这道题要求记录下标，故这个方法行不通。

python代码如下：

 1   def twoSum(self, num, target):
 2     d = {}
 3     lenNum = len(num)
 4     ans = []
 5     for i in xrange(lenNum):
 6       d[num[i]] = [i+1] if num[i] not in d else d[num[i]] + [i+1]
 7     for i in num:
 8       res = target - i
 9       if res in d:
10         if res == i:
11           if len(d[i]) == 2: return (d[i][0], d[i][1])
12         else:
13           return (d[i][0], d[res][0]) if d[i][0] < d[res][0] else (d[res][0], d[i][0])

 2. 3Sum(https://oj.leetcode.com/problems/3sum/)

(1)对num进行排序

(2)取一个元素a, 从a的下一个元素到数组的末尾，采用两端夹逼的方法，找到b, c符合b+c=target-a

(3)此过程中注意去重

(4)时间复杂度O(n^2)

 1   def threeSum(self, num):
 2     res = []
 3     lenNum = len(num)
 4     if lenNum < 2: return res
 5     num.sort()
 6     for i in xrange(lenNum-2):
 7       a = num[i]
 8       if i > 0 and a == num[i-1]:
 9         continue
10       left = i+1; right = lenNum-1
11       while left < right:
12         b = num[left]
13         c = num[right]
14         if b + c == -a:
15           res.append([a, b, c])
16           while left < right:
17             left += 1
18             right -= 1
19             if num[left] != b or num[right] != c:
20               break
21         elif b + c < -a:
22           while left < right:
23             left += 1
24             if num[left] != b:
25               break
26         else:
27           while left < right:
28             right -= 1
29             if num[right] != c:
30               break
31     return res

3. 3Sum Closest(https://oj.leetcode.com/problems/3sum-closest/)

(1)对num进行排序

(2)先取定a, 采用两端夹逼的方法取b, c， 计算sum=a+b+c, 维护mindiff=target-sum，若碰到sum=target，则立即返回0。

(3)时间复杂度O(n^2)

  def threeSumClosest(self, num, target):
    res = []
    num.sort()
    mindiff = 100000
    lenNum = len(num)
    for i in xrange(lenNum-2):
      a = num[i]
      left = i+1
      right = lenNum-1
      while left < right:
        b = num[left]
        c = num[right]
        sum = a + b + c
        diff = abs(target - sum)
        if diff < mindiff: mindiff = diff; res = sum
        if sum < target:
          left += 1
        elif sum > target:
          right -= 1
        else:
          return sum
    return res

 4. 4Sum(https://oj.leetcode.com/problems/4sum/)

(1)对num进行排序

(2)使用dict,存储dict[num[i]+num[j]]=([i,j]), key为num[i]+num[j], value为[[i,j]]即下标。

(3)两层循环, p, q, num[p] < num[q], 则在dict中找key为target-num[p]-num[q]的value，且q < i

(4)时间复杂度O(n^2)

 1   def fourSum(self, num, target):
 2     num.sort()
 3     res = set()
 4     lenNum = len(num)
 5     dict = {}
 6     for i in xrange(lenNum):
 7       for j in xrange(i+1, lenNum):
 8         key = num[i] + num[j]
 9         if key not in dict:
10           dict[key] = [(i, j)]
11         else:
12           dict[key].append((i, j))
13 
14     for p in xrange(lenNum):
15       for q in xrange(p+1, lenNum):
16         key = target - num[p] - num[q]
17         if key in dict:
18           for value in dict[key]:
19             if q < value[0]:
20               res.add((num[p], num[q], num[value[0]], num[value[1]]))
21     return [list(i) for i in res]

 

posted @ 2014-10-23 23:35  darlwen  阅读(202)  评论(0)  编辑  收藏  举报