704. Binary Search

Given a sorted (in ascending order) integer array nums of n elements and a target value, write a function to search target in nums. If target exists, then return its index, otherwise return -1.


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

Note:

  1. You may assume that all elements in nums are unique.
  2. n will be in the range [1, 10000].
  3. The value of each element in nums will be in the range [-9999, 9999].
class Solution {
    public int search(int[] nums, int target) {
        List<Integer> list = new ArrayList();
        for(int i: nums) list.add(i);
        return list.indexOf(target);
    }
}

 二分法爬

class Solution {
    public int search(int[] nums, int target) {
        int left = 0, right = nums.length - 1;
        while(left <= right){
            int mid = left + (right - left)/2;
            if(target == nums[mid]) return mid;
            else if(nums[mid] > target) right = mid - 1;
            else left = mid + 1;
        }
        return -1;
    }
}

开玩笑的,这题得用left <= right, 不然会出现很多鸡掰的情况

原文地址:https://www.cnblogs.com/wentiliangkaihua/p/12892837.html