栏目分类:
子分类:
返回
终身学习网用户登录
快速导航关闭
当前搜索
当前分类
子分类
实用工具
热门搜索
终身学习网 > IT > 软件开发 > 后端开发 > Java

力扣算法 Java 刷题笔记【数组篇(一) 二分搜索】hot100(一)二分查找、搜索插入位置、在排序数组中查找元素的第一个和最后一个位置 3

Java 更新时间:发布时间: 百科书网 趣学号

文章目录

1. 二分查找(简单)2. 搜索插入位置(简单)3. 在排序数组中查 找元素的第一个和最后一个位置(中等)

1. 二分查找(简单)

地址: https://leetcode-cn.com/problems/binary-search/
2022/01/30
做题反思:

class Solution {
    public int search(int[] nums, int target) {
        int left = 0, right = nums.length - 1;
        while (left <= right) {
            int mid = left + (right - left) / 2;
            if (nums[mid] == target) {
                return mid;
            }
            if (nums[mid] < target) {
                left = mid + 1;
            }
            if (nums[mid] > target) {
                right = mid - 1;
            }
        }
        return -1;
    }
}

2. 搜索插入位置(简单)

地址: https://leetcode-cn.com/problems/search-insert-position/
2022/01/30
做题反思:

class Solution {
    public int searchInsert(int[] nums, int target) {
        int left = 0, right = nums.length - 1;
        while (left <= right) {
            int mid = left + (right - left) / 2;
            if (nums[mid] == target) {
                right = mid - 1;
            }
            if (nums[mid] < target) {
                left = mid + 1;
            }
            if (nums[mid] > target) {
                right = mid - 1;
            }
        }
        return left;
    }
}

3. 在排序数组中查 找元素的第一个和最后一个位置(中等)

地址: https://leetcode-cn.com/problems/find-first-and-last-position-of-element-in-sorted-array/
2022/01/30
做题反思:

class Solution {
    public int[] searchRange(int[] nums, int target) {
        return new int[]{leftSearch(nums, target), rightSearch(nums, target)};
    }

    int leftSearch(int[] nums, int target) {
        int left = 0, right = nums.length - 1;
        while (left <= right) {
            int mid = left + (right - left) / 2;
            if (nums[mid] == target) {
                right = mid - 1;
            }
            if (nums[mid] < target) {
                left = mid + 1;
            }
            if (nums[mid] > target) {
                right = mid - 1;
            }
        }
        if (left >= nums.length || nums[left] != target) {
            return -1;
        }
        return left;
    }

    int rightSearch(int[] nums, int target) {
        int left = 0, right = nums.length - 1;
        while (left <= right) {
            int mid = left + (right - left) / 2;
            if (nums[mid] == target) {
                left = mid + 1;
            }
            if (nums[mid] < target) {
                left = mid + 1;
            }
            if (nums[mid] > target) {
                right = mid - 1;
            }
        }
        if (right < 0 || nums[right] != target) {
            return -1;
        }
        return right;
    }
}

转载请注明:文章转载自 www.051e.com
本文地址:http://www.051e.com/it/728057.html
我们一直用心在做
关于我们 文章归档 网站地图 联系我们

版权所有 ©2023-2025 051e.com

ICP备案号:京ICP备12030808号