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