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Given a target number and an integer array A sorted in ascending order, find the index i in A such that A[i] is closest to the given target.
Return -1 if there is no element in the array.
Given [1, 2, 3] and target = 2, return 1.
Given [1, 4, 6] and target = 3, return 1.
Given [1, 4, 6] and target = 5, return 1 or 2.
Given [1, 3, 3, 4] and target = 2, return 0 or 1 or 2.
There can be duplicate elements in the array, and we can return any of the indices with same value.
1 class Solution { 2 public: 3 /** 4 * @param A an integer array sorted in ascending order 5 * @param target an integer 6 * @return an integer 7 */ 8 int closestNumber(vector<int>& A, int target) { 9 if (A.size() == 0) { 10 return -1; 11 } 12 13 int b = 0, e = A.size() - 1; 14 while (b + 1 < e) { 15 int mid = b + (e - b) / 2; 16 if (A[mid] == target) { 17 return mid; 18 } else if (A[mid] < target) { 19 b = mid; 20 } else { 21 e = mid; 22 } 23 } 24 if (abs(A[b] - target) < abs(A[e] - target)) { 25 return b; 26 } else { 27 return e; 28 } 29 } 30 };
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原文地址：http://www.cnblogs.com/whlsai/p/5141308.html
