标签:div indexof 情况 code may ret otherwise lis src
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 innumsand its index is 4
Example 2:
Input:nums= [-1,0,3,5,9,12],target= 2 Output: -1 Explanation: 2 does not exist innumsso return -1
Note:
nums are unique.n will be in the range [1, 10000].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
标签:div indexof 情况 code may ret otherwise lis src
原文地址:https://www.cnblogs.com/wentiliangkaihua/p/12892837.html