简单二分(折半查找) 二分算法复杂度:O(log2n) 1 int BinarySearch(const int test[], int len, int target) 2 { 3 int left = 0, right = len - 1, mid; 4 while (left <= right) 5 { 6 mid = (left + right) / 2; 7 if (test[mid] == target) 8 { 9 return mid; 10 } 11 else if (test[mid] > target) 12 { 13 right ...