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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

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
    }
}

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