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‎docs/0000-00-note.adoc‎

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* [ ] 通过 xref:2560-house-robber-iv.adoc[2560. House Robber IV] 发现，竟然还有「最大化最小值」或「最小化最大值」的二分答案问题。
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* [ ] KMP 算法
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* [ ] 可以使用类库的，却总想自己写代码实现！必然字符串比较！
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* [ ] 在 https://leetcode.cn/problems/search-insert-position/solutions/2023391/er-fen-cha-zhao-zong-shi-xie-bu-dui-yi-g-nq23/[35. 搜索插入位置 - 二分查找总是写不对？一个视频讲透！^] 中，作者说二分查找“在找到 target 就立刻返回”，这种写法适用范围很窄，多思考体会一下。
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== 感慨
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‎docs/0000-01-modified-binary-search.adoc‎

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== 疑问点
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. 二分搜索怎样可以确定地取左右端点？
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. 在 https://leetcode.cn/problems/search-insert-position/solutions/2023391/er-fen-cha-zhao-zong-shi-xie-bu-dui-yi-g-nq23/[35. 搜索插入位置 - 二分查找总是写不对？一个视频讲透！^] 中，作者说“在找到 target 就立刻返回”，这种写法适用范围很窄，多思考体会一下。
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== 经典题目
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‎docs/0035-search-insert-position.adoc‎

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[#0035-search-insert-position]
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= 35. Search Insert Position
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= 35. 搜索插入位置
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{leetcode}/problems/search-insert-position/[LeetCode - Search Insert Position^]
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https://leetcode.cn/problems/search-insert-position/[LeetCode - 35. 搜索插入位置^]
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Given a sorted array and a target value, return the index if the target is found. If not, return the index where it would be if it were inserted in order.
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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。
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You may assume no duplicates in the array.
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请必须使用时间复杂度为 stem:[O(log n)] 的算法。
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*Example 1:*
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*示例 1:*
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[subs="verbatim,quotes,macros"]
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----
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*Input:* [1,3,5,6], 5
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*Output:* 2
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----
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....
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输入: nums = [1,3,5,6], target = 5
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输出: 2
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....
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*Example 2:*
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*示例 2:*
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[subs="verbatim,quotes,macros"]
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----
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*Input:* [1,3,5,6], 2
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*Output:* 1
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----
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....
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输入: nums = [1,3,5,6], target = 2
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输出: 1
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....
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*Example 3:*
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*示例 3:*
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[subs="verbatim,quotes,macros"]
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----
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*Input:* [1,3,5,6], 7
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*Output:* 4
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----
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....
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输入: nums = [1,3,5,6], target = 7
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输出: 4
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....
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*Example 4:*
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*提示:*
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* `1 \<= nums.length \<= 10^4^`
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* `-10^4^ \<= nums[i] \<= 10^4^`
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* `nums` 为**无重复元素**的**升序**排列数组
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* `-10^4^ \<= target \<= 10^4^`
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[subs="verbatim,quotes,macros"]
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----
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*Input:* [1,3,5,6], 0
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*Output:* 0
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----
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== 思路分析
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image::images/modified-binary-search.jpeg[{image_attr}]
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[[src-0035]]
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[tabs]
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====
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一刷::
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+
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--
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[{java_src_attr}]
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----
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include::{sourcedir}/_0035_SearchInsertPosition.java[tag=answer]
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----
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--
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二刷::
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--
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[{java_src_attr}]
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----
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include::{sourcedir}/_0035_SearchInsertPosition_2.java[tag=answer]
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----
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--
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====
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== 参考资料
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. https://leetcode.cn/problems/search-insert-position/solutions/2023391/er-fen-cha-zhao-zong-shi-xie-bu-dui-yi-g-nq23/[35. 搜索插入位置 - 二分查找总是写不对？一个视频讲透！^]
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. https://leetcode.cn/problems/search-insert-position/solutions/10969/te-bie-hao-yong-de-er-fen-cha-fa-fa-mo-ban-python/[35. 搜索插入位置 - 尽可能讲到大家对于二分查找可能遇到的问题，结尾有题单（Java）^]
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. https://leetcode.cn/problems/search-insert-position/solutions/536357/yi-wen-dai-ni-gao-ding-er-fen-cha-zhao-j-69ao/[35. 搜索插入位置 - 一文带你搞定二分查找及其多个变种！^]

‎logbook/202503.adoc‎

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|{doc_base_url}/0039-combination-sum.adoc[题解]
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|✅ 回溯。使用回溯解法，要注意剪枝技巧。这道题还是一个“完全背包”问题，也可以尝试一下完全背包的解法。
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|{counter:codes2503}
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|{leetcode_base_url}/search-insert-position/[35. 搜索插入位置^]
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|{doc_base_url}/0035-search-insert-position.adoc[题解]
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|✅ 二分查找。主要学习二分查找的各种变种解法。解答中提到“在找到 target 就立刻返回”，这种写法适用范围很窄，多思考体会一下。
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|===
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截止目前，本轮练习一共完成 {codes2503} 道题。

‎src/main/java/com/diguage/algo/leetcode/_0035_SearchInsertPosition.java‎

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* @since 2018-09-16 21:32
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*/
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public class _0035_SearchInsertPosition {
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// tag::answer[]
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// tag::answer[]
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/**
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* @author D瓜哥 · https://www.diguage.com
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* @since 2018-09-16 21:32
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*/
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public static int searchInsert(int[] nums, int target) {
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if (null == nums || nums.length == 0) {
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return 0;
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return result;
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}
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// end::answer[]
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// end::answer[]
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public static void main(String[] args) {

‎src/main/java/com/diguage/algo/leetcode/_0035_SearchInsertPosition_2.java‎

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package com.diguage.algo.leetcode;
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public class _0035_SearchInsertPosition_2 {
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// tag::answer[]
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/**
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* @author D瓜哥 · https://www.diguage.com
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* @since 2025-11-26 21:03:14
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*/
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public int searchInsert(int[] nums, int target) {
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int left = 0, right = nums.length - 1;
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while (left <= right) {
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// 循环不变量：
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// nums[left-1] < target
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// nums[right+1] >= target
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int mid = left + (right - left) / 2;
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int num = nums[mid];
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if (num < target) {
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left = mid + 1;
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} else {
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right = mid - 1;
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}
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}
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return right + 1; // 这里使用 left 也可以
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}
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// end::answer[]
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static void main() {
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new _0035_SearchInsertPosition_2()
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.searchInsert(new int[]{1, 3, 5, 6}, 2);
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}
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}

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