List-binarysearch
二分查找
leetcode.704
- 链接https://leetcode.cn/problems/...
解题方法：二分查找模板（有序数组）
bool check(int x) // 检查x是否满足某种性质 int bsearch_1(int l, int r){ while (l < r){ int mid = l + r >> 1; if (check(mid)) r = mid; else l = mid + 1; } return l; } int bsearch_2(int l, int r){ while (l < r){ int mid = l + r + 1>> 1; if (check(mid)) l = mid; else r = mid - 1; } return l; }leetcode解题代码
class Solution { public: int search(vector& nums, int target) { int l = 0, r = nums.size() - 1; while (l < r){ int mid = (l + r) / 2; if (nums[mid] >= target) r = mid; else l = mid + 1; } if (nums[l] == target) return l; return -1; } }; - ACM模式调试
输入
第一行输入两个数n，target
n表示数组中数的个数，target表示目标值
第二行表示数组
5 9
-1 0 3 5 9 12输出
4调试代码
#include
#include
using namespace std;
int main(){
int n, target;
vector nums(6);
cin >> n >> target;
for (int i = 0; i < n; i ++) cin >> nums[i];
int l = 0, r = nums.size() - 1;
while (l < r){
int mid = (l + r) / 2;
if (nums[mid] >= target) r = mid;
else l = mid + 1;
}
if (nums[l] == target) cout << l;
else cout << -1;
return 0;
} leetcode.35
- 链接https://leetcode.cn/problems/...
- 解题方法：注意这里与上一题的区别在于需要特判，如果目标值大于数组最后一位则返回数组长度
leetcode解题代码
class Solution { public: int searchInsert(vector& nums, int target) { int n = nums.size(); if (target > nums[n - 1]) return n; int l = 0, r = n - 1; while (l < r){ int mid = (l + r) / 2; if (nums[mid] >= target) r = mid; else l = mid + 1; } return l; } }; - ACM模式调试 和上题类似
leetcode.69
- 链接https://leetcode.cn/problems/...
- 解题方法：本题题意类似于找到最大的y使得y^2<=x，所以使用第二个模板
leetcode解题代码
class Solution { public: int mySqrt(int x) { long l = 0, r = x;// long防止越界 while (l < r){ int mid = l + r + 1 >> 1; if (mid <= x / mid) l = mid;// mid * mid会越界 else r = mid - 1; } return r; } };- ACM模式调试
输入一个数x
4输出
2调试代码
#include
using namespace std;
int main(){
int x;
cin >> x;
long l = 0, r = x;// long防止越界
while (l < r){
int mid = l + r + 1 >> 1;
if (mid <= x / mid) l = mid;// mid * mid会越界
else r = mid - 1;
}
cout << r << endl;
return 0;
} leetcode.367
- 链接https://leetcode.cn/problems/...
- 解题方法：与上题类似，注意数组越界
leetcode解题代码
class Solution { public: bool isPerfectSquare(int num) { long l = 1, r = num; while (l < r){ int mid = l + r + 1 >> 1; if (mid <= num / mid) l = mid; else r = mid - 1; } return r * r == num; } };- ACM模式调试代码 和上题类似
leetcode.34
- 链接https://leetcode.cn/problems/...
- 解题方法：找到数组的二段性，利用模板一找到左侧第一个target的下标，利用模板二找到右侧最后一个target的下标
对于第一个target，[5, 7, 7, 8]，nums[mid] < target
对于最后一个target，[8, 10]，nums[mid] > target leetcode解题代码
class Solution { public: vectorsearchRange(vector & nums, int target) { int l = 0, r = nums.size() - 1; while (l < r) { int mid = (l + r) / 2; if (nums[mid] < target) l = mid + 1; else r = mid; } if (nums[l] != target) return {-1, -1}; int L = l; l = 0, r = nums.size() - 1; while (l < r) { int mid = (l + r + 1) / 2; if (nums[mid] > target) r = mid - 1; else l = mid; } return {L, r}; } }; - ACM模式调试
输入
第一行输入两个数n，target
n表示数组中数的个数，target表示目标值
第二行表示数组
6 8
5 7 7 8 8 10输出
3 4 调试代码
#include
#include
using namespace std;
int main(){
int n, target;
vector nums(6);
cin >> n >> target;
for (int i = 0; i < n; i ++) cin >> nums[i];
vector res;
int l = 0, r = nums.size() - 1;
while (l < r)
{
int mid = (l + r) / 2;
if (nums[mid] < target) l = mid + 1;
else r = mid;
}
if (nums[l] != target) cout << -1 << -1 << endl;
res.push_back(l);
int L = l;
l = 0, r = nums.size() - 1;
while (l < r)
{
int mid = (l + r + 1) / 2;
if (nums[mid] > target) r = mid - 1;
else l = mid;
}
res.push_back(l);
for (auto c: res){
cout << c << ' ';
}
return 0;
} 解题参考：https://www.acwing.com/
刷题顺序：https://www.programmercarl.com/