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340、搜索插入位置

圣杰 2022-01-04 阅读 29

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
示例 4:

输入: nums = [1,3,5,6], target = 0
输出: 0
示例 5:

输入: nums = [1], target = 0
输出: 0
 

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

public class SearchInsertPosition {

    /**

     * @param args

     */

    public static void main(String[] args) {

        SearchInsertPosition sip = new SearchInsertPosition();

        int[] nums1 = new int[] {135,6};

       System.out.println(sip.searchInsert(nums1, 5));

        int[] nums2 = new int[] {1};

        System.out.println(sip.searchInsert(nums2, 0));

        int[] nums3 = new int[] {135,6};

        System.out.println(sip.searchInsertLogN(nums3, 2));

    }

    //nums = [1,3,5,6], target = 5,输出2

    //nums = [1], target = 0,输出0

    public int searchInsert(int[] nums, int target) {

        int insertIndex = 0;

        for (int i = 0; i < nums.length; i++) {

            if (target < nums[i]) {

                return i;

            }else if(target == nums[i]) {

                return i;

            }else if(target> nums[i]) {

                insertIndex++;

            }

        }

        return insertIndex;

    }

     

    //nums = [1,3,6], target = 5,输出2

    public int searchInsertLogN(int[] nums, int target) {

        int left = 0;

        //注意:-1

        int right=nums.length-1;

        //注意:<=

        while(left<=right) {

            int mid  = (left+right)/2;

            if(nums[mid] == target) {

                return mid;

            }else if(nums[mid] < target) {

                left=mid+1;

            }else {

                right=mid-1;

            }

        }

        return left;

    }

}

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