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https://github.com/halfrost/LeetCode-Go.git
synced 2025-07-05 00:25:22 +08:00
添加 problem 34
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@ -0,0 +1,78 @@
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package leetcode
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func searchRange(nums []int, target int) []int {
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return []int{searchFirstEqualElement(nums, target), searchLastEqualElement(nums, target)}
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}
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// 二分查找第一个与 target 相等的元素,时间复杂度 O(logn)
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func searchFirstEqualElement(nums []int, target int) int {
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low, high := 0, len(nums)-1
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for low <= high {
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mid := low + ((high - low) >> 1)
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if nums[mid] > target {
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high = mid - 1
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} else if nums[mid] < target {
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low = mid + 1
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} else {
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if (mid == 0) || (nums[mid-1] != target) { // 找到第一个与 target 相等的元素
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return mid
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}
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high = mid - 1
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}
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}
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return -1
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}
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// 二分查找最后一个与 target 相等的元素,时间复杂度 O(logn)
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func searchLastEqualElement(nums []int, target int) int {
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low, high := 0, len(nums)-1
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for low <= high {
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mid := low + ((high - low) >> 1)
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if nums[mid] > target {
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high = mid - 1
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} else if nums[mid] < target {
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low = mid + 1
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} else {
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if (mid == len(nums)-1) || (nums[mid+1] != target) { // 找到最后一个与 target 相等的元素
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return mid
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}
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low = mid + 1
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}
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}
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return -1
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}
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// 二分查找第一个大于等于 target 的元素,时间复杂度 O(logn)
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func searchFirstGreaterElement(nums []int, target int) int {
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low, high := 0, len(nums)-1
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for low <= high {
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mid := low + ((high - low) >> 1)
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if nums[mid] >= target {
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if (mid == 0) || (nums[mid-1] < target) { // 找到第一个大于等于 target 的元素
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return mid
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}
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high = mid - 1
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} else {
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low = mid + 1
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}
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}
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return -1
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}
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// 二分查找最后一个小于等于 target 的元素,时间复杂度 O(logn)
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func searchLastLessElement(nums []int, target int) int {
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low, high := 0, len(nums)-1
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for low <= high {
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mid := low + ((high - low) >> 1)
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if nums[mid] <= target {
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if (mid == len(nums)-1) || (nums[mid+1] > target) { // 找到最后一个小于等于 target 的元素
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return mid
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}
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low = mid + 1
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} else {
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high = mid - 1
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}
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}
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return -1
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}
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@ -0,0 +1,47 @@
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package leetcode
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import (
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"fmt"
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"testing"
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)
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type question34 struct {
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para34
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ans34
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}
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// para 是参数
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// one 代表第一个参数
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type para34 struct {
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nums []int
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target int
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}
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// ans 是答案
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// one 代表第一个答案
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type ans34 struct {
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one []int
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}
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func Test_Problem34(t *testing.T) {
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qs := []question34{
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question34{
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para34{[]int{5, 7, 7, 8, 8, 10}, 8},
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ans34{[]int{3, 4}},
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},
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question34{
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para34{[]int{5, 7, 7, 8, 8, 10}, 6},
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ans34{[]int{-1, -1}},
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},
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}
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fmt.Printf("------------------------Leetcode Problem 34------------------------\n")
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for _, q := range qs {
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_, p := q.ans34, q.para34
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fmt.Printf("【input】:%v 【output】:%v\n", p, searchRange(p.nums, p.target))
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}
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fmt.Printf("\n\n\n")
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}
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@ -0,0 +1,36 @@
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# [34. Find First and Last Position of Element in Sorted Array](https://leetcode.com/problems/find-first-and-last-position-of-element-in-sorted-array/)
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## 题目:
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Given an array of integers `nums` sorted in ascending order, find the starting and ending position of a given `target` value.
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Your algorithm's runtime complexity must be in the order of *O*(log *n*).
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If the target is not found in the array, return `[-1, -1]`.
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**Example 1:**
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Input: nums = [5,7,7,8,8,10], target = 8
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Output: [3,4]
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**Example 2:**
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Input: nums = [5,7,7,8,8,10], target = 6
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Output: [-1,-1]
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## 题目大意
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给定一个按照升序排列的整数数组 nums,和一个目标值 target。找出给定目标值在数组中的开始位置和结束位置。你的算法时间复杂度必须是 O(log n) 级别。如果数组中不存在目标值,返回 [-1, -1]。
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## 解题思路
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- 给出一个有序数组 `nums` 和一个数 `target`,要求在数组中找到第一个和这个元素相等的元素下标,最后一个和这个元素相等的元素下标。
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- 这一题是经典的二分搜索变种题。二分搜索有 4 大基础变种题:
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1. 查找第一个值等于给定值的元素
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2. 查找最后一个值等于给定值的元素
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3. 查找第一个大于等于给定值的元素
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4. 查找最后一个小于等于给定值的元素
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这一题的解题思路可以分别利用变种 1 和变种 2 的解法就可以做出此题。或者用一次变种 1 的方法,然后循环往后找到最后一个与给定值相等的元素。不过后者这种方法可能会使时间复杂度下降到 O(n),因为有可能数组中 n 个元素都和给定元素相同。(4 大基础变种的实现见代码)
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