Given a sorted array and a target value, return the index if the target is found. If not, return the index where it would be if it were inserted in order.
You may assume NO duplicates in the array.
注意：要找的数比数组里的数都大，因此插入的位置为A.length.
1 public class Solution { 2 /** 3 * param A : an integer sorted array 4 * param target : an integer to be inserted 5 * return : an integer 6 */ 7 public int searchInsert(int[] A, int target) { 8 if (A == null || A.length == 0) { 9 return 0; 10 } 11 int begin = 0; 12 int end = A.length - 1; 13 while (begin + 1 < end) { 14 int mid = begin + (end - begin) / 2; 15 if (A[mid] == target) { 16 return mid; 17 } 18 if (A[mid] > target) { 19 end = mid; 20 } else { 21 begin = mid; 22 } 23 } 24 if (A[begin] >= target) { 25 return begin; 26 } else if (A[end] >= target){ 27 return end; 28 } else { 29 return end + 1; 30 } 31 } 32 }
原文：http://www.cnblogs.com/FLAGyuri/p/5370616.html