[leetcode]搜索插入位置题解

问题说明

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

解决方案

package leetcode.editor.cn;
/**
 * 搜索插入位置
 * @date 2022-02-11 14:19:13
 * @author 沙漠渔
 */
public class 搜索插入位置{
static
//leetcode submit region begin(Prohibit modification and deletion)
class Solution {
    public int searchInsert(int[] nums, int target) {
        int left = 0, right = nums.length - 1;
        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;
    }
}
//leetcode submit region end(Prohibit modification and deletion)

    static class ListNode {
        int val;
        ListNode next;
        public ListNode(int val) { this.val = val; }
        public ListNode(int val, ListNode next) { this.val = val; this.next = next; }
    }
	
	/**
     * 最终测试主方法
     */
	public static void main(String[] args) {
        Solution solution = new Solution();
        System.out.println(solution.searchInsert(new int[]{1,2,3,4},3));
    }
}