给定一个排序数组和一个目标值，在数组中找到目标值，并返回其索引。如果目标值不存在于数组中，返回它将会被按顺序插入的位置。
请必须使用时间复杂度为 O (log n) 的算法。
示例 1:
输入: nums = [1,3,5,6], target = 5 | |
输出: 2 |
示例 2:
输入: nums = [1,3,5,6], target = 2 | |
输出: 1 |
示例 3:
输入: nums = [1,3,5,6], target = 7 | |
输出: 4 |
提示:
1 <= nums.length <= 104 | |
-104 <= nums[i] <= 104 | |
nums 为 无重复元素 的 升序 排列数组 | |
-104 <= target <= 104 |
题解
/** | |
* 二分查找法 | |
* | |
*/ | |
public int searchInsert(int[] nums, int target) { | |
int length = nums.length; | |
int left = 0; | |
int right = length - 1; | |
while (left <= right){ | |
int mid = (left + right) / 2; | |
if (target > nums[mid]){ | |
left = mid + 1; | |
}else { | |
right = mid - 1; | |
} | |
} | |
return left; | |
} |
递归
public int searchInsert(int[] nums, int target) { | |
return binarySearch(nums, target, 0, nums.length - 1); | |
} | |
private int binarySearch(int[] nums, int target ,int left, int right){ | |
if (left > right) return left; | |
int mid = (left + right) /2; | |
if (target > nums[mid]){ | |
return binarySearch(nums, target, mid + 1, right); | |
}else if (target < nums[mid]){ | |
return binarySearch(nums, target, left, mid - 1); | |
}else { | |
return mid; | |
} | |
} |
来源：力扣（LeetCode）
链接：https://leetcode.cn/problems/search-insert-position
