剑指offer——和为s的两个数字VS和为s的连续正数序列

来源：互联网发布：人工智能综合报告编辑：程序博客网时间：2024/05/16 12:14

两种方法都类似于快排的变形。

#include <iostream>#include <string>using namespace std;bool FindNumbersWithSum(int data[],int length,int sum,int *num1,int *num2){bool found=false;if (length<1||num1==NULL||num2==NULL){return found;}int ahead=length-1;int behind=0;while(ahead>behind){long curSum=data[ahead]+data[behind];if (curSum==sum){*num1=data[behind];*num2=data[ahead];found=true;break;}else if(curSum>sum){ahead--;}else{behind++;}}return found;}void PrintContinuousSequence(int small,int big){for (int i=small;i<=big;i++){cout<<i<<" ";}cout<<endl;}void FindContinuousSequence(int sum){if (sum<3){return;}int small=1;int big=2;int middle=(1+sum)/2;int curSum=small+big;while(small<middle){if (curSum==sum){PrintContinuousSequence(small,big);}while(curSum>sum&&small<middle){curSum-=small;small++;if (curSum==sum){PrintContinuousSequence(small,big);}}big++;curSum+=big;}}int main(){int str[]={1,2,3,4,5,6,7,8,9,10};int m,n;FindNumbersWithSum(str,10,13,&m,&n);cout<<m<<n<<endl;FindContinuousSequence(100);return 0;}

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