题目如下：
下面是我写的一些解答，，有更好的做法或者写法欢迎指正。
(1)
#include<iostream>#include<algorithm>#include<vector>#include<cstdlib>using namespace std;/* * (1) A数组是否有重复元素或者说前后值相差为1 * (2) A数组是否有序 * */void slove(int A[],int len){//assert(A!=NULL);int lastValue=A[0];sort(A,A+len);for(int i=0;i<len;i++){if(i==0){cout<<0<<"-"<<(A[0]>1?A[0]-1:0)<<endl;}else{if(A[i]-1<=lastValue){lastValue=A[i];continue;}else{cout<<lastValue+1<<"-"<<A[i]-1<<endl;lastValue=A[i];}}}if(lastValue<100){cout<<lastValue+1<<"-"<<100<<endl;}}int main(){int testA[]={4,4,4,27};slove(testA,4);}(2) 递归写法#include<iostream>//int sumValue=0;using namespace std;#define len 7void sum(int A[],int min,int max,int i,int &sumValue){if(i<=len){if(A[i]>=min&&A[i]<max){sumValue+=A[i];sum(A,min,A[i],i*2,sumValue);sum(A,A[i],max,i*2+1,sumValue);}else if(A[i]<min){//右子树sum(A,min,max,i*2+1,sumValue);}else if(A[i]>=max){//左子树sum(A,min,max,i*2,sumValue);}}}int main(){/*5 */\ *38 */ \/ \ *14 7 10 * */ int A[]={0,5,3,8,1,4,7,10}; int sumValue=0; sum(A,3,7,1,sumValue); cout<<"sumValue="<<sumValue<<endl;}(4) 二分查找的变种，稍稍改进一下二分搜索时间复杂度log(n)#include<iostream>using namespace std;int binarySearch(int A[],int l,int r,int s){if(l<=r){int mid=l+(r-l)/2;if(A[mid]<s){//右子树查找if(mid+1<=r&&A[mid]<s&&A[mid+1]>=s) //相邻的加判断return A[mid];else if(mid+1>r&&A[mid]<s) return A[mid]; //处理一下特殊情况elsereturn binarySearch(A,mid+1,r,s);}else{if(mid-1>=l&&A[mid-1]<s&&A[mid]>=s)return A[mid-1];elsereturn binarySearch(A,l,mid-1,s);}}return -1; //找不到就返回-1}int main(){ int A[]={5,6,7,8}; int val; if((val=binarySearch(A,0,3,10))!=-1){cout<<val<<endl; }else{cout<<"在数组中没有找到合适的值"<<endl; }}
总在盼望未来,愿你的人生美开
