Kotlin Kata - Binary Search
Given an array of integers nums which is sorted in ascending order, and an integer target, write a function to search target in nums. If target exists, then return its index. Otherwise, return -1.
You must write an algorithm with O(log n) runtime complexity.
Example 1:
Input: nums = [-1,0,3,5,9,12], target = 9 Output: 4 Explanation: 9 exists in nums and its index is 4
Example 2:
Input: nums = [-1,0,3,5,9,12], target = 2 Output: -1 Explanation: 2 does not exist in nums so return -1
class Solution {
fun search(nums: IntArray, target: Int): Int {
}
}
解答
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nums 已經排序過
利用 Binary Search 的邏輯
我們可以每次去除一半的選項
讓時間複雜度限制在 O(log n)
實作的部分
我們可以用一個 while 迴圈搜尋
class Solution {
fun search(nums: IntArray, target: Int): Int {
var left = 0
var right = nums.size - 1
while (left <= right) {
val pivot = left + (right - left) / 2
when {
nums[pivot] == target -> return pivot
nums[pivot] > target -> right = pivot - 1
nums[pivot] < target -> left = pivot + 1
}
}
return -1
}
}