php数组的一些常见操作汇总

代码如下:

// 三个数组的共同元素-只找最小的
void FindCommonElements(int a[], int b[], int c[], int x, int y, int z)
{
for(int i = 0, j = 0, k = 0; i {
if(a[i] {
i++ ;
}
else // a[i] >= b[j]
{
if(b[j] {
j++ ;
}
else // b[j] >= c[k]
{
if(c[k] {
k++ ;
}
else // c[k] >= a[i]
{
cout <return ;
}
}
}
}

cout <<"Not found!" <}

代码如下:

// Find the unique common element in 3 arrays
// O(NlogN)
int UniqueCommonItem(int *a, int *b, int *c, int n)
{
// sort array a
qsort(a, n, sizeof(int), compare) ; // NlogN

// sort array b
qsort(b, n, sizeof(int), compare) ; // NlogN

// for each element in array c, do a binary search in a and b
// This is up to a complexity of N*2*logN
for (int i = 0; i {
if(BinarySearch(a, n, c[i]) && BinarySearch(b, n, c[i]))
return c[i] ;
}

return - 1 ; // not found
}

代码如下:

// Find the unique common element in 3 arrays
// O(NlogN)
int UniqueCommonItem1(int *a, int *b, int *c, int n)
{
// sort array a
qsort(a, n, sizeof(int), compare) ; // NlogN

// Space for time
bool *bb = new bool[n] ;
memset(bb, 0, n) ;

bool *bc = new bool[n] ;
memset(bb, 0, n) ;

// for each element in b, do a BS in a and mark all the common element
for (int i = 0; i {
if(BinarySearch(a, n, b[i]))
bb[i] = true ;
}

// for each element in c, do a BS only if b[i] is true
for (int i = 0; i {
if(b[i] && BinarySearch(a, n, c[i]))
return c[i] ;
}

return - 1 ; // not found
}

代码如下:

// Determine whether a contains value k
bool BinarySearch(int *a, int n, int k)
{
int left = 0 ;
int right = n - 1 ;
while (left <= right)
{
int mid = (left + right) ;

if(a[mid] left = mid + 1 ;
if(a[mid] == k)
return true ;
else
right = mid - 1 ;
}

return false ;
}

// Compare function for qsort
int compare(const void* a, const void* b)
{
return *(int*)a - *(int*)b ;
}

代码如下:

// 找出满足给定和的数对
void FixedSum(int* a, int* b, int n, int d)
{
for (int i = 0, j = n - 1; i = 0)
{
if (a[i] + b[j] ++i ;
else if (a[i] + b[j] == d)
{
cout <++i ;
--j ;
}
else // a[i] + b[j] > d
--j ;
}
}

代码如下:

// 子数组的最大和
int Sum(int* a, int n)
{
int curSum = 0;
int maxSum = 0;
for (int i = 0; i {
if (curSum + a[i] <0)
curSum = 0;
else
{
curSum += a[i] ;
maxSum = max(maxSum, curSum);
}
}
return maxSum;
}

代码如下:

// 子数组的最大乘积
int MaxProduct(int *a, int n)
{
int maxProduct = 1; // max positive product at current position
int minProduct = 1; // min negative product at current position
int r = 1; // result, max multiplication totally

for (int i = 0; i {
if (a[i] > 0)
{
maxProduct *= a[i];
minProduct = min(minProduct * a[i], 1);
}
else if (a[i] == 0)
{
maxProduct = 1;
minProduct = 1;
}
else // a[i] <0
{
int temp = maxProduct;
maxProduct = max(minProduct * a[i], 1);
minProduct = temp * a[i];
}

r = max(r, maxProduct);
}

return r;
}

代码如下:

// 将buffer中start和end之间的元素逆序
void Reverse( int buffer[], int start, int end )
{
while ( start {
int temp = buffer[ start ] ;
buffer[ start++ ] = buffer[ end ] ;
buffer[ end-- ] = temp ;
}
}

// 将含有n个元素的数组buffer右移k位
void Shift( int buffer[], int n, int k )
{
k %= n ;

Reverse( buffer, 0, n - k - 1) ;
Reverse( buffer, n - k, n - 1 ) ;
Reverse( buffer, 0, n - 1 ) ;
}

代码如下:

// 字符串逆序
void Reverse(char *a, int n)
{
int left = 0;
int right = n - 1;

while (left {
char temp = a[left] ;
a[left++] = a[right] ;
a[right--] = temp ;
}
}

代码如下:

// n选m的所有组合
int buffer[100] ;

void PrintArray(int *a, int n)
{
for (int i = 0; i cout <cout <}

bool IsValid(int lastIndex, int value)
{
for (int i = 0; i {
if (buffer[i] >= value)
return false;
}
return true;
}

void Select(int t, int n, int m)
{
if (t == m)
PrintArray(buffer, m);
else
{
for (int i = 1; i <= n; i++)
{
buffer[t] = i;
if (IsValid(t, i))
Select(t + 1, n, m);
}
}
}

代码如下:

// 合并两个有序数组
void Merge(int *a, int *b, int *c, int n)
{
int i = 0 ;
int j = 0 ;
int k = 0 ;

while (i {
if (a[i] {
c[k++] = a[i] ;
++i ;
}
else if (a[i] == b[j])// 如果a和b元素相等，则插入二者皆可，这里插入a
{
c[k++] = a[i] ;
++i ;
++j ;
}
else // a[i] > b[j] // 如果b中元素小，则插入b中元素到c
{
c[k++] = b[j] ;
++j ;
}
}

if (i == n) // 若a遍历完毕，处理b中剩下的元素
{
for (int m = j; m c[k++] = b[m] ;
}
else//j == n, 若b遍历完毕，处理a中剩下的元素
{
for (int m = i; m c[k++] = a[m] ;
}
}

代码如下:

void Arrange(int* a, int n)
{
int k = n - 1 ;
for (int i = n - 1; i >= 0; --i)
{
if (a[i] != 0)
{
if (a[k] == 0)
{
a[k] = a[i] ;
a[i] = 0 ;
}
--k ;
}
}
}