刷题顺序按照代码随想录建议
题目描述
英文版描述
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
numsare unique. numsis sorted in ascending order.
英文版地址
中文版描述
给定一个 n 个元素有序的（升序）整型数组 nums 和一个目标值 target ，写一个函数搜索 nums 中的 target，如果目标值存在返回下标，否则返回 -1。
示例 1:
输入: nums = [-1,0,3,5,9,12], target = 9
输出: 4
解释: 9 出现在 nums 中并且下标为 4
示例 2:
输入: nums = [-1,0,3,5,9,12], target = 2
输出: -1
解释: 2 不存在 nums 中因此返回 -1
提示：
- 你可以假设
nums中的所有元素是不重复的。 n将在[1, 10000]之间。nums的每个元素都将在[-9999, 9999]之间。
中文版地址
解题方法
俺这版
class Solution {
public int search(int[] nums, int target) {
int leftPointer = 0;
int rightPointer = nums.length - 1;
while (leftPointer <= rightPointer) {
int midIndex = leftPointer +
(int) Math.ceil((rightPointer - leftPointer) / 2.0);
if (midIndex == leftPointer) {
if (nums[midIndex] == target) {
return midIndex;
}
return -1;
}
if (nums[midIndex] > target) {
rightPointer = midIndex - 1;
} else {
leftPointer = midIndex;
}
}
return -1;
}
}
复杂度分析
- 时间复杂度：O(logN)，N是数组长度
- 空间复杂度：O(1)
官方版
class Solution {
public int search(int[] nums, int target) {
int left = 0, right = nums.length - 1;
while (left <= right) {
int mid = (right - left) / 2 + left;
int num = nums[mid];
if (num == target) {
return mid;
} else if (num > target) {
right = mid - 1;
} else {
left = mid + 1;
}
}
return -1;
}
}
复杂度分析
- 时间复杂度：O(logN)，N是数组长度
- 空间复杂度：O(1)
