binary search【leetcode34,35,74,240】
一、模板
题目：
A = [1,2,3,3,4]
target = 3<<#key points:
#start + 2 < end
#start + (end - start)/2
#A[mid] ==.<.>
#A[start] A[end] ? target
#适用范围：可以转换为first/last position of 的题目start = 0
end = len(A)-1
while(start+1 target
end = mid
if A[start] == target: #若找last position，两个if换一下
return start
if A[end] == target:
return end
return -1 二、【leetcode34】search for a range
Given an array of integers sorted in ascending order, find the starting and ending position of a given target value.
Your algorithm's runtime complexity must be in the order of O(log n).
If the target is not found in the array, return [-1, -1].
For example,
Given [5, 7, 7, 8, 8, 10] and target value 8,
return [3, 4]
class Solution(object):
def searchRange(self, nums, target):
"""
:type nums: List[int]
:type target: int
:rtype: List[int]
"""
start = 0
end = len(nums)-1
result = [-1,-1]
if nums == []: #注意特殊情况
return result
while (start+1 < end):
mid = start + (end-start)/2
if nums[mid] == target:
end = mid
elif nums[mid] < target:
start = mid
elif nums[mid] > target:
end = mid
if nums[start] == target:
result[0] = start
elif nums[end] == target:
result[0] = end
else:
return result
start = 0
end = len(nums)-1
while (start+1 < end):
mid = start + (end-start)/2
if nums[mid] == target:
start = mid
elif nums[mid] < target:
start = mid
elif nums[mid] > target:
end = mid
if nums[end] == target:
result[1] = end
elif nums[start] == target:
result[1] = start
else:
return result
return result模板应用：分别寻找 first position和last position，用两次模板
三、【leetcode35】search insert position
Given a sorted array and a target value, return the index if the target is found. If not, return the index where it would be if it were inserted in order.
You may assume no duplicates in the array.
Here are few examples.[1,3,5,6], 5 → 2[1,3,5,6], 2 → 1[1,3,5,6], 7 → 4[1,3,5,6], 0 → 0
class Solution(object):
def searchInsert(self, nums, target):
"""
:type nums: List[int]
:type target: int
:rtype: int
"""
if nums == []:
return 0
start = 0
end = len(nums)-1
while start + 1 < end:
mid = start + (end - start)/2
if nums[mid] >= target:
end = mid
elif nums[mid] < target:
start = mid
if nums[start] >= target:
return start
elif nums[end] >= target:
return end
else:
return len(nums) #注意考虑边界四、【leetcode74】search A 2D Matrix
Write an efficient algorithm that searches for a value in an m x n matrix. This matrix has the following properties:
- Integers in each row are sorted from left to right.
- The first integer of each row is greater than the last integer of the previous row.
For example,
Consider the following matrix:
[ [1, 3, 5, 7], [10, 11, 16, 20], [23, 30, 34, 50] ]
Given target = 3, return true.
class Solution(object):
def searchMatrix(self, matrix, target):
"""
:type matrix: List[List[int]]
:type target: int
:rtype: bool
"""
if matrix == []:
return False
if matrix == [[]]:
return False
m,n = len(matrix), len(matrix[0])
start = 0
end = m*n-1
while start + 1 < end:
mid = start + (end - start)/2
x,y = mid/n, mid%n #注意是m还是n
if matrix[x][y] < target:
start = mid
else:
end = mid
x,y = start/n, start%n
if matrix[x][y] == target:
return True
x,y = end/n, end%n
if matrix[x][y] == target:
return True
return False如果用for将matrix遍历，再对每个matrix[i]进行二分法查找，时间复杂度为O（n2）；
如果在确定哪一行时使用找到matrix[i][0]或matrix[i][-1]，会有边界问题，较为复杂；
方法应为end = m*n -1，用x=mid/n.y=mid%n,确定元素位置的方法使用二分法。
五、【leetcode240】search a 2D maxtrix 2
1、
Write an efficient algorithm that searches for a value in an m x n matrix. This matrix has the following properties:
- Integers in each row are sorted in ascending from left to right.
- Integers in each column are sorted in ascending from top to bottom.
For example,
Consider the following matrix:
[ [1, 4, 7, 11, 15], [2, 5, 8, 12, 19], [3, 6, 9, 16, 22], [10, 13, 14, 17, 24], [18, 21, 23, 26, 30] ]
Given target = 5, return true.
Given target = 20, return false.
从左下角查起，如果matrix[x][y] < target，删去第x列，反之，删去第y行，如果相等，return true。这样的时间复杂度为m+n
3、
class Solution(object):
def searchMatrix(self, matrix, target):
"""
:type matrix: List[List[int]]
:type target: int
:rtype: bool
"""
if matrix == []:
return False
if matrix == [[]]:
return False
m,n = len(matrix)-1,0
while n < len(matrix[0]) and m >= 0:
if matrix[m][n] < target:
n += 1
elif matrix[m][n] > target:
m -= 1
elif matrix[m][n] == target:
return True
return False