[Binary Search] Leetcode 35, 74
阿新 • • 發佈:2018-09-18
none win div des search arr left 二分 sel
35. Search Insert Position
Description
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.
Example 1:
Input: [1,3,5,6], 5 Output: 2
Example 2:
Input: [1,3,5,6], 2 Output: 1
Example 3:
Input: [1,3,5,6], 7 Output: 4
Example 4:
Input: [1,3,5,6], 0 Output: 0
Solution
二分法查找
1 class Solution: 2 def searchInsert(self, nums, target): 3 """ 4 :type nums: List[int] 5 :type target: int 6 :rtype: int 7 """ 8 l, r = 0, len(nums)9 10 while l < r: 11 m = (l + r) // 2 12 if nums[m] == target: 13 return m 14 elif nums[m] > target: 15 r = m 16 else: 17 l = m + 1 18 return l
74. Search a 2D Matrix
Description
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.
Example 1:
Input: matrix = [ [1, 3, 5, 7], [10, 11, 16, 20], [23, 30, 34, 50] ] target = 3 Output: true
Example 2:
Input: matrix = [ [1, 3, 5, 7], [10, 11, 16, 20], [23, 30, 34, 50] ] target = 13 Output: false
Solution
Approach 1: 二分查找
先按行二分確定目標所在行;再在該行中確定元素位置。
1 class Solution: 2 def searchMatrix(self, matrix, target): 3 """ 4 :type matrix: List[List[int]] 5 :type target: int 6 :rtype: bool 7 """ 8 if not matrix or not matrix[0]: return False 9 m, n = len(matrix), len(matrix[0]) 10 11 if target < matrix[0][0] or target > matrix[m - 1][n - 1]: 12 return False 13 14 start, end = 0, m 15 while start < end: 16 midrow = (start + end) // 2 17 if matrix[midrow][0] == target: 18 return True 19 elif target < matrix[midrow][0]: 20 end = midrow 21 else: start = midrow + 1 22 row = start - 1 23 24 l, r = 0, n 25 26 while l < r: 27 midcol = (l + r) // 2 28 if matrix[row][midcol] == target: return True 29 elif matrix[row][midcol] < target: 30 l = midcol + 1 31 else: r = midcol 32 return False
Beats: 41.43%
Runtime: 48ms
Approach 2: 從右上到左下查找
若當前 > target, 則向前一列查找 => 則矩陣後幾列均不用再考慮;
若當前 < target, 則向下一行查找 => 則矩陣前幾行均不用再考慮。
1 class Solution: 2 def searchMatrix(self, matrix, target): 3 """ 4 :type matrix: List[List[int]] 5 :type target: int 6 :rtype: bool 7 """ 8 if not matrix or not matrix[0]: return False 9 10 rows, cols = len(matrix), len(matrix[0]) 11 row, col = 0, cols - 1 12 while True: 13 if row < rows and col >= 0: 14 if matrix[row][col] == target: 15 return True 16 elif matrix[row][col] < target: 17 row += 1 18 else: col -= 1 19 else: return False
Beats: 41.67%
Runtime: 48ms
[Binary Search] Leetcode 35, 74