快速排序、求旋转数组最小数字

#include <iostream>#include <stack>#include <queue>#include <algorithm>using namespace std;// 快速排序1void quickSort( int a[], int l, int r ){if( l >= r ){return;}int i = l; int j = r;int mid = ( l + r ) / 2;int tmp = a[mid];a[mid] = a[l];a[l] = tmp;int value = a[l];while( i < j ){while( i < j && a[j] >= value ) j--;a[i] = a[j];while( i < j && a[i] <= value ) i++;a[j] = a[i];}a[i] = value;quickSort( a, l, i - 1 );quickSort( a, i + 1, r );return;}// 快速排序2，容易理解的快排void quickSort1( int a[], int start, int end ){if( start >= end ){return;}int mid = ( start + end ) / 2;int tmp = a[mid];a[mid] = a[end];a[end] = tmp;int idx = start - 1;for( int i = start; i < end; i++ ){if( a[i] < a[end] )   // a[end]作为pivot {idx++;if( idx != i ){int tmp = a[idx];a[idx] = a[i];a[i] = tmp;}}}idx++;tmp = a[idx];a[idx] = a[end];a[end] = tmp;quickSort1( a, start, idx - 1 );quickSort1( a, idx + 1, end );return;}void print( int a[], int len ){for( int i = 0; i < len; i++ ){cout << a[i] << ' ';}cout << endl;return;}// 求旋转数组的最小数字 递归实现int midRotatedArray( int a[], int start, int end ){int min = -1;if( NULL == a ){min = -1;return min;}if( start >= end ){min = a[start];return min;}int mid = ( start + end ) / 2;//if( a[mid] <= a[ ( mid + 1 ) % int right = mid + 1 > end ? start : mid + 1;int left = mid - 1 < start ? end: mid - 1;if( a[mid] <= a[right] && a[mid] <= a[left] )  {if( !( a[mid] == a[right] && a[mid] == a[left] ) ) // for { 2 3 3 3 1 }{min = a[mid];//cout << start << ' ' << mid << ' ' << end << endl;return min;}}int minLeft = midRotatedArray( a, start, mid - 1 );int minRight = midRotatedArray( a, mid + 1, end );return minLeft > minRight ? minRight : minLeft;}// 求旋转数组的最小数字 非递归实现int midRotatedArray_noRecur( int a[], int start, int end ){int min = -1;if( NULL == a ){min = -1;return min;}if( start >= end ){min = a[start];return min;}int mid = ( start + end ) / 2;while( a[start] >= a[end] ){mid = ( start + end ) / 2;int right = mid + 1 > end ? start : mid + 1;int left = mid - 1 < start ? end : mid - 1;if( a[mid] <= a[right] && a[mid] <= a[left] ){if( !( a[mid] == a[right] && a[mid] == a[left] ) ){min = a[mid];//cout << start << ' ' << mid << ' ' << end << endl;return min;}}if( a[start] == a[mid] && a[mid] == a[end] ) // 顺序查找, for : { 1 0 1 1 1 } and  { 1 1 1 0 1 }{min = a[start];for( int i = start + 1; i <= end; i++ ){if( min > a[i] ){min = a[i];}}return min;}if( start != mid && a[start] >= a[mid] ){end = mid - 1;}else {start = mid + 1;}}min = a[start];return min;}// 总结： 先写测试用例，否则很多考虑不周int main(){int ret = 0;int data = -2;data = data << 2;cout << data << endl;data = -2;data = data >> 2;cout << data << endl;        // 针对求旋转数组的最小数字的测试用例int b0[] = { 3, 4, 5, 6, 1 };int b1[] = { 2, 2, 3, 0, 1 };int b2[] = { 1, 2, 3, 4, 5 };int b3[] = { 2, 2, 2, 2, 2 };int b4[] = { 1, 1, 1, 0, 1 };int b5[] = { 1, 0, 1, 1, 1 };int b6[] = { 2, 1 };int b7[] = { 3 };int *b8 = NULL;cout << midRotatedArray( b7, 0, 0 ) << endl;cout << midRotatedArray_noRecur( b7, 0, 0 ) << endl;int a1[] = { 1, 2, 3, 4, 5 };int a2[] = { 5, 4, 3, 2, 1 };int a3[] = { 5, 7, 1, 3, 6 };int a4[] = { 5, 4, 4, 4, 1 };int a5[] = { 4, 4, 4, 4, 4 };quickSort1( a1, 0, 4 );quickSort1( a2, 0, 4 );quickSort1( a3, 0, 4 );quickSort1( a4, 0, 4 );quickSort1( a5, 0, 4 );print( a1, 5 );print( a2, 5 );print( a3, 5 );print( a4, 5 );print( a5, 5 );return ret;}