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LeetCode--Search for a Range
Given a sorted array of integers, find the starting and ending position of a given target value.
Your algorithm‘s runtime complexity must be in the order of O(log n).
If the target is not found in the array, return [-1, -1]
.
For example,
Given [5, 7, 7, 8, 8, 10]
and target value 8,
return [3, 4]
.
class Solution { public: vector<int> searchRange(int A[], int n, int target) { vector<int> res; if(n == 0 ) return res; int low = search_low(A,0,n-1,target); res.push_back(low); if(low == -1) res.push_back(low); else { int high = search_high(A,0,n-1,target); res.push_back(high); } return res; } int search_low(int* a, int low, int high, int target) { int loc = -1; while(low <= high) { int mid = (low+high)/2; if(a[mid] == target) { loc = mid; high = mid-1; } else if(a[mid] > target) high = mid-1; else low = mid+1; } return loc; } int search_high(int* a, int low, int high, int target) { int loc = -1; while(low <= high) { int mid = (low+high)/2; if(a[mid] == target) { loc = mid; low = mid+1; } else if(a[mid] > target) high = mid-1; else low = mid+1; } return loc; } };
LeetCode--Search for a Range
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