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.
分析
使用binary Search 找到可以插入 target 的 position, 例如是 i, 那么，从i（包括i）到后面，都是大于等于target的数字
1 如果 i == 0, return i，因为A[i]一定是最接近 target的数字，后面的数字都大于A[i]
2 如果 i > =，那么需要考虑A[i] 和 A[i - 1] 哪个更接近 target，就返回哪一个
1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27 | public class Solution { /** * @param A an integer array sorted in ascending order * @param target an integer * @return an integer */ public int closestNumber(int[] A, int target) { // Write your code here if(A == null || A.length == 0) return -1; int left = 0, right = A.length - 1, mid; while(left < right){ mid = left + (right - left) / 2; if(A[mid] < target){ left = mid + 1; } else{ right = mid; } } // when left == right, this is the first position that target can be insert if(right > 0 && (A[right] - target) > (target - A[right - 1])) return right - 1; else return right; }} |