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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(vector<int>& nums, int target) { vector<int> res; res.push_back(bsarch_left(nums, target)); res.push_back(bsarch_right(nums, target)); return res; } int bsarch_left(vector<int>& nums, int target) { int i = 0; int j = nums.size() - 1; while (i < j) { int mid = i + (j-i) / 2; if (nums[mid] < target) { i = mid + 1; } else { j = mid; } } if (target == nums[i]) { return i; } else { return -1; } } int bsarch_right(vector<int>& nums, int target) { int i = 0; int j = nums.size() - 1; while (i < j) { int mid = i + (j-i + 1) / 2; if (nums[mid] <= target) { i = mid; } else { j = mid - 1; } } if (target == nums[i]) { return i; } else { return -1; } } };
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原文地址：http://www.cnblogs.com/SpeakSoftlyLove/p/5184624.html
