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.
[1,3,5,6], 5 → 2
[1,3,5,6], 2 → 1
[1,3,5,6], 7 → 4
[1,3,5,6], 0 → 0
public class Solution { /** * param A : an integer sorted array * param target : an integer to be inserted * return : an integer */ public int searchInsert(int[] A, int target) { // write your code here if(A == null || A.length == 0) return 0; int left = 0; int right = A.length - 1; if(target <= A[left]) return 0; if(target > A[right]) return A.length; while(left < right - 1){ int mid = left + (right - left) / 2; if(A[mid] == target) return mid; else if(A[mid] < target) left = mid; else right = mid - 1; } if(A[left] < target) return left + 1; return left; } }
lintcode-easy-Search Insert Position
原文：http://www.cnblogs.com/goblinengineer/p/5249268.html