Divide and conquer strategy exercises for algorithm design and analysis
- Divide and conquer strategy to solve problem one: search in half
- Divide and conquer strategy to solve the problem two: binary search technology application one
- Divide and conquer strategy to solve problem three: binary search technology application two
- Divide and conquer strategy to solve problem four: mode and multiplicity
- Divide and conquer strategy to solve problem five: interval merger
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Divide and conquer strategy to solve problem one: search in half
Problem Description
Given that n elements a[0:n-1] have been sorted in ascending order, now we need to find a specific element x among these n elements.
problem analysis
Start searching from the middle of the element, and search for x by halving continuously, and the sub-problems for searching in half are independent of each other.
Algorithm implementation
//折半查找递归实现
int binarySearch(int a[],int left,int right,int x){
if(left<=right){
int middle = (left + right) / 2;
if(x == a[middle]) return middle;
else if(x < a[middle]) return binarySearch(a,left,middle-1,x);
else return binarySearch(a,middle+1,right,x);
}
return -1; //没找到返回-1
}
Divide and conquer strategy to solve the problem two: binary search technology application one
Problem Description
Given a number of integers, ask if the sum of a pair of numbers is equal to the given number.
Input:
4
2 5 1 4
6
Output
1 5
Explanation: If there are multiple pairs of numbers that meet the requirements, output the pair with the smallest number
problem analysis
- Sort the array from small to large first
- Use binary search to find a value corresponding to the former
- Stop searching when found, otherwise continue
Algorithm implementation
//折半查找递归实现
int binarySearch(int a[],int left,int right,int x){
if(left<=right){
int middle = (left + right) / 2;
if(x == a[middle]) return middle;
else if(x < a[middle]) return binarySearch(a,left,middle-1,x);
else return binarySearch(a,middle+1,right,x);
}
return -1; //没找到返回-1
}
void SumNumberSearch(int a[],int length,int sum){
for(int i=0;i<length;i++){
if(binarySearch(a,i+1,length-1,sum-a[i])){
cout<<a[i]<<" "<<sum-a[i];
break;
}
}
}
Divide and conquer strategy to solve problem three: binary search technology application two
Problem Description
Input n numbers, output them from small to large, and output the repeated numbers only once.
Input:
5
2 4 4 5 1
Output:
1 2 4 5
problem analysis
The easiest way is to use set, use the nature of set, here we do not use this method, using binary search technology
- First sort the array from small to large
- Starting from the middle number, record the first and last occurrence of the number respectively
- Analyze the subsequence on the left in the same way
- Output the number
- Analyze the right subsequence in the same way
Algorithm implementation
void lineSearch(int a[],int left,int right){
int number,middle,l,r,i;
if(left>right) return;
middle = (left + right)/2;
number = a[middle];
i = middle - 1;
while(a[i]==number && i>=left) i--;
l = i;
i = middle + 1;
while(a[i]==number && i<=right) i++;
r = i;
lineSearch(a,left,l);
cout<<number<<" ";
lineSearch(a,r,right);
}
Divide and conquer strategy to solve problem four: mode and multiplicity
Problem Description
For a given multiple number set S consisting of n natural numbers, program to calculate the mode and multiplicity of S
problem analysis
- Use array storage
- Sort the array first
- Record the index of the middle number, the number of occurrences of the middle number, and the position of the first occurrence of the middle number
- Compare the multiplicity of the intermediate number with the multiplicity of the previous record, if it is greater than the multiplicity of the previous record, update num and sum
- If the largest multiplicity is less than the number of digits to the right of the middle number, then recursive analysis to the right
- If the largest multiplicity is less than the number of digits to the left of the middle number, then recursively analyze to the left
Algorithm implementation
#include <iostream>
#include <algorithm>
#define N 100
using namespace std;
int num = 0; //存储众数
int sum = 0; //存储重数
/*
统计中间数出现的数量
*/
int count(int[],int,int);
/*
找到中间数第一次出现的位置
*/
int start(int[],int,int);
/*
找众数和其重数
*/
void modeAndMultiplicity(int[],int,int);
int main()
{
int a[N],n;
cout<<"请输入数组元素数量：";
cin>>n;
cout<<"请输入数组元素：";
for(int i=0;i<n;i++)
cin>>a[i];
sort(a,a+n);
modeAndMultiplicity(a,0,n-1);
cout<<"众数"<<num<<"的重数为"<<sum<<endl;
return 0;
}
/*
统计中间数出现的数量 实现
*/
int count(int a[],int front,int rear){
int i = 0; //计数器
int mid = a[(front+rear)/2];
for(int j=front;j<=rear;j++){
if(a[j] == mid){
i++;
}
}
return i;
}
/*
找到中间数第一次出现的位置 实现
*/
int start(int a[],int front,int rear){
int x = 0;
int mid = a[(front+rear)/2];
for(int i=front;i <= rear;i++){
if(a[i] == mid){
x = i;
break;
}
}
return x;
}
/*
找众数和其重数 实现
*/
void modeAndMultiplicity(int a[],int front,int rear){
int mNum = (front+rear)/2; //当前中间数的下标
int mSum = count(a,front,rear); //当前中间数的重数
int mLeft = start(a,front,rear); //当前中间数第一次出现的位置
if(mSum > sum) {
//重数大则替换众数和其重数
sum = mSum;
num = a[mNum];
}
if(rear-(mLeft+mSum)+1 > sum){
//右边数量大于重数 则向右找
modeAndMultiplicity(a,mLeft+mSum,rear);
}
if(mLeft > sum){
//左边数量大于重数 则向左找
modeAndMultiplicity(a,front,mLeft-1);
}
}
Divide and conquer strategy to solve problem five: interval merger
Problem Description
Given n closed intervals [ai; bi], where i=1,2,...,n. Any two adjacent or intersecting closed intervals can be merged into one closed interval. For example, [1;2] and [2;3] can be combined into [1;3], [1;3] and [2;4] can be combined into [1;4], which is [1;2] and [ 3;4] Can't merge.
Determine whether these intervals can be finally merged into a closed interval, if so, output the closed interval, otherwise output no.
problem analysis
- Define interval structure for storage and define comparison function
- After sorting the input interval groups, divide and conquer on the left and right sides until it is reduced to the merge problem between two small areas
- If it cannot be combined, the program outputs no and exits directly
- If possible, merge the results after the end of the divide and conquer to get the final interval
Algorithm implementation
#include <iostream>
#include <algorithm>
//定义区间数量
#define N 100
using namespace std;
//定义区间结构体
struct Interval{
int left,right;
};
//定义区间比较函数
bool compareInterval(Interval,Interval);
//区间合并
Interval intervalMerge(Interval[],int,int);
int main()
{
Interval interval[N];
int n;
cout<<"请输入区间数量：";
cin>>n;
for(int i=0;i<n;i++)
cin>>interval[i].left>>interval[i].right;
sort(interval,interval+n,compareInterval);
Interval tempInterval = intervalMerge(interval,0,n-1);
cout<<tempInterval.left<<" "<<tempInterval.right<<endl;
return 0;
}
bool compareInterval(Interval mLeft,Interval mRight){
if(mLeft.left < mRight.left)
return true;
return false;
}
Interval intervalMerge(Interval internal[],int left,int right){
if(left == right)
return internal[left];
Interval tempInterval1 = intervalMerge(internal,left,(left+right)/2),
tempInterval2 = intervalMerge(internal,(left+right)/2+1,right);
if(tempInterval1.right >= tempInterval2.left){
//符合合并条件 进行合并
Interval tempInterval;
tempInterval.left = tempInterval1.left;
tempInterval.right = tempInterval1.right>=tempInterval2.right
?tempInterval1.right
:tempInterval2.right;
return tempInterval;
}else{
//否则退出
cout<<"no"<<endl;
exit(0);
}
}
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