二分查找的各种条件

int firstEqual(int *a, int beg, int end, int x) {// a[i-1] < x <= a[i]
int s = beg - 1;
int t = end + 1;
while ( s + 1 < t)
{
int mid = s + (t - s)/2;
if (x > a[mid])
s = mid;
else
t = mid;
}

//要返回后面的坐标, 所以要对后面的坐标进行判断
if (t == end + 1 || a[t] != x)
return -1;
return t;

}

// a[i - 1]<= x < a[i]， 对前面的坐标进行判断
int lastEqual(int *a, int beg, int end, int x) {
int s = beg -1;
int t = end +1;
while (s +1 < t) {
int mid = s + (t - s) /2;
if (x >= a[mid])
s = mid;
else
t = mid;
}

if (s == beg - 1 || a[s] != x)
return -1;
return s;
}

// a[i - 1] <= x <a[i], 判断后面的坐标
int firstBigger(int *a, int beg, int end, int x) {//找到比x大的第一个数
int s = beg - 1;
int t = end + 1;

while (s + 1 < t) {
int mid = s + (t - s)/2;
if (x >= a[mid])
s = mid;
else
t= mid;
}

if (t == end + 1 || a[t] <= x)
return -1;
return t;
}

//a[i-1] < x <= a[i], 判断前面的坐标
int lastLess(int *a, int beg, int end, int x) {//找到比x小的最后一个数
int s = beg - 1;
int t = end + 1;

while (s +1 < t)
{
int mid = s + (t - s)/2;
if (x > a[mid])
s = mid;
else
t = mid;
}
if (s == beg - 1 || a[s] >= x)
return -1;
return s;
}

int main()
{
int a[] = {1,2,2,2,2,2,5,5,5,5,5};//10
int b[] = {1,3};
int c[] = {1,2,5,6,6,6,6};//6
int d[] = {3,4,6,6,7,7,9,10,10};//8

cout<<firstBigger(a,0,10,2)<<" "<<firstBigger(a,0,10,5)<<" "<<firstBigger(a,0,10,3)<<" "<<firstBigger(a,0,10,1)<<endl;
cout<<firstBigger(b,0,1,1)<<" "<<firstBigger(b,0,1,2)<<endl;
cout<<firstBigger(c,0,6,2)<<" "<<firstBigger(c,0,6,5)<<" "<<firstBigger(c,0,6,3)<<" "<<firstBigger(c,0,6,6)<<endl;
cout<<firstBigger(d,0,8,3)<<" "<<firstBigger(d,0,8,4)<<" "<<firstBigger(d,0,8,6)<<" "<<firstBigger(d,0,8,7)<<" "<<firstBigger(d,0,8,9)<<" "<<firstBigger(d,0,8,10)<<" "<<firstBigger(d,0,8,2)<<endl;
cout<<endl;

cout<<lastLess(a,0,10,2)<<" "<<lastLess(a,0,10,5)<<" "<<lastLess(a,0,10,3)<<" "<<lastLess(a,0,10,1)<<endl;
cout<<lastLess(b,0,1,1)<<" "<<lastLess(b,0,1,2)<<endl;
cout<<lastLess(c,0,6,2)<<" "<<lastLess(c,0,6,5)<<" "<<lastLess(c,0,6,3)<<" "<<lastLess(c,0,6,6)<<endl;
cout<<lastLess(d,0,8,3)<<" "<<lastLess(d,0,8,4)<<" "<<lastLess(d,0,8,6)<<" "<<lastLess(d,0,8,7)<<" "<<lastLess(d,0,8,9)<<" "<<lastLess(d,0,8,10)<<" "<<lastLess(d,0,8,2)<<endl;

return 0;
}

所有条件的二分查找只有两种方式

1.

mid mid

a[i-1]<x<=a[i]

s t

因为t的那一边有一个等号，所以当 a[mid] >= x时，t = mid

1.1 返回大于等于x的第一个 return t

1.2 返回小于x的最后一个 return s

2.

mid mid

a[i-1]<=x<a[i]

s t

因为s的那一边有等号，所以当a[mid] <=x时，s = mid

2.1 返回小于等于x的最后一个 return s

2.2 返回大于x的第一个 return t