35. Search Insert Position
题目:
Given a sorted array and a target value, return the index if the target is found. If not, return the index where it would be if it were inserted in order.
You may assume no duplicates in the array.
Here are few examples. [1,3,5,6]
, 5 → 2 [1,3,5,6]
, 2 → 1 [1,3,5,6]
, 7 → 4 [1,3,5,6]
, 0 → 0
代码:
第一反应,遍历了一遍:
public int searchInsert(int[] nums, int target) {
int result =0;
for (int i = 0 ; i < nums.length ; i++)
{
if (target <= nums[i])
{
result = i;
break;
}
else
{
result=i+1;
}
}
return result;
}
网上查了一下,发现二分法,算法复杂度会减少不少,尤其在较大的数据量:
int searchInsert_mid(int[] nums, int target) {
int low = 0, high = nums.length-1;
while(low <= high) {
int mid = (low+high)>>1;
if(nums[mid] == target)
return mid;
else if(nums[mid] > target)
high = mid-1;
else
low = mid+1;
}
if(high < 0) return 0;
if(low >= nums.length) return nums.length;
return low;
}
【推荐】编程新体验,更懂你的AI,立即体验豆包MarsCode编程助手
【推荐】凌霞软件回馈社区,博客园 & 1Panel & Halo 联合会员上线
【推荐】抖音旗下AI助手豆包,你的智能百科全书,全免费不限次数
【推荐】轻量又高性能的 SSH 工具 IShell:AI 加持,快人一步