• LeetCode-704. 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

    Constraints:

    • 1 <= nums.length <= 10^4
    • -10^4 < nums[i], target < 10^4
    • All the integers in nums are unique.
    • nums is sorted in ascending order.
    class Solution {
      //nums用到的数组。   target所选择的索引目标
        public int search(int[] nums, int target) {
          //对nums数组进行遍历
            for(int i =0;i < nums.length;i++){
              //如果nums遍历之中的索引i的值 正好等于 target
                if(nums[i] == target){
                  //那么返回i
                    return i;
                }
                
            }
          //否则返回-1
            return -1;
        }
    }
    
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  • 原文地址:https://www.cnblogs.com/pengcode/p/15311824.html
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