930. Binary Subarrays With Sum
题目描述
数组A仅包含0和1，有多少和为S的非空子数组。
例子
Input: A = [1,0,1,0,1], S = 2
Output: 4
Explanation:
The 4 subarrays are bolded below:
[1,0,1]
[1,0,1]
[0,1,0,1]
[1,0,1]
思想
非空子数组，起始位置为i，截止位置为j，暴力先求前缀数组和，再取差值作为A[i:j]的和，时间复杂度O(n^2)。
因为数组只包含0和1，所以可以观察出数组的前缀和是一递增序列。需要在该序列中找到两数差值为S，想到2Sum。
建一个字典，记录当前不同前缀和的数量。
dic[i]表示和为i的前缀个数，sum[i]表示到i的前缀和。
若当前坐标是j，则目标是计算j之前共有多少个前缀和等于sum[j] - S，即dic[sum[j] - S]。
解法1
暴力，TLE
class Solution(object):
def numSubarraysWithSum(self, A, S):
"""
:type A: List[int]
:type S: int
:rtype: int
"""
n = len(A)
summ= [0] * (n+1)
for i in range(1, n+1):
summ[i] = summ[i-1] + A[i-1]
cnt = 0
for i in range(n+1):
for j in range(i+1, n+1):
if summ[j] - summ[i] == S:
cnt += 1
return cnt
（改进）
class Solution(object):
def numSubarraysWithSum(self, A, S):
"""
:type A: List[int]
:type S: int
:rtype: int
"""
cnt = summ = 0
dic = {0:1}
for num in A:
summ += num
if summ - S in dic:
cnt += dic[summ-S]
if summ in dic:
dic[summ] += 1
else:
dic[summ] = 1
return cnt
简洁
class Solution(object):
def numSubarraysWithSum(self, A, S):
"""
:type A: List[int]
:type S: int
:rtype: int
"""
cnt = summ = 0
dic = {0:1}
for num in A:
summ += num
cnt += dic.get(summ - S, 0)
dic[summ] = dic.get(summ, 0) + 1
return cnt
228. Summary Ranges - Easy
题目描述
给定无重复的有序整数数组，将其拆成连续的范围。
例子
Example 1:
Input: [0,1,2,4,5,7]
Output: [“0->2”,“4->5”,“7”]
Explanation: 0,1,2 form a continuous range; 4,5 form a continuous range.
Example 2:
Input: [0,2,3,4,6,8,9]
Output: [“0”,“2->4”,“6”,“8->9”]
Explanation: 2,3,4 form a continuous range; 8,9 form a continuous range.
思想
因为数组已排好序，所以遍历一遍判断并存储结果即可。
Trick：’->’.join(map(str, p))
解法
最笨拙直接的方法
class Solution(object):
def summaryRanges(self, nums):
"""
:type nums: List[int]
:rtype: List[str]
"""
if not nums:
return []
res = []
temp = [nums[0], nums[0]]
for num in nums[1:]:
if num == temp[-1] + 1:
temp[-1] = num
else:
if temp[0] == temp[1]:
res.append(str(temp[0]))
else:
res.append('%s->%s'%(temp[0], temp[1]))
temp = [num, num]
if temp[0] == temp[1]:
return res + [str(temp[0])]
return res + ['%s->%s'%(temp[0], temp[1])]
简化代码：最后一起添加’->’
class Solution(object):
def summaryRanges(self, nums):
"""
:type nums: List[int]
:rtype: List[str]
"""
res = []
for num in nums:
if res and num == res[-1][-1] + 1:
res[-1] = [res[-1][0], num]
else:
res.append([num])
return ['->'.join(map(str, p)) for p in res]