转载自:https://www.jianshu.com/p/0f823fbd4d20
二分查找很好写,却很难写对,出错原因主要集中在判定条件和边界值的选择上,很容易就会导致越界或者死循环的情况。
下面对二分查找及其变形进行总结:
1. 最基本的二分查找
int a[100005]; int find1(int target, int l,int r)//l,r是查找的左右区间 { int left = l, right = r, mid; while (left <= right) { mid = left + (right - left) / 2; if (a[mid] == target) return mid; else if (a[mid] > target) right = mid - 1; else left = mid + 1; } return -1; }
其中,有几个要注意的点:
- 循环的判定条件是:
low <= high
- 为了防止数值溢出,
mid = low + (high - low)/2
- 当
A[mid]
不等于target
时,high = mid - 1
或low = mid + 1
2. 当目标值有多个的时候,返回从左往右的第一个目标值。(目标值区域的左边界/查找与目标值相等的第一个位置/查找第一个不小于目标值数的位置)
eg:
A = [1,3,3,5, 7 ,7,7,7,8,14,14]
target = 7
return 4
int a[100005]; int find2(int target,int l,int r) { int left = l, right = r, mid; while (left <= right) { mid = left + (right - left) / 2; if (target <= a[mid]) right = mid - 1; else left = mid + 1; } if (left <= r && a[left] == target) return left; else return -1; }
3. 当目标值有多个的时候,返回从左往右的最后一个目标值。(查找目标值区域的右边界/查找与目标值相等的最后一个位置/查找最后一个不大于目标值数的位置)
eg:
A = [1,3,3,5,7,7,7, 7 ,8,14,14]
target = 7
return 7
int a[100005]; int find3(int target, int l,int r) { int left = l, right = r, mid; while (left <= right) { mid = left + (right - left) / 2; if (target >= a[mid]) left = mid + 1; else right = mid - 1; } if (right >= 0 && a[right] == target) return right; else return -1; }