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.
Here are few examples.[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 { public int searchInsert(int[] nums, int target) { //二分法的变形，没有找到时，返回的是应该插入的下标 return find(nums,target,0,nums.length-1); } public int find(int []nums,int target,int start,int end){ if(start>end) return start;//二分法唯一的不同 int mid=(start+end)/2; if(nums[mid]==target) return mid; if(nums[mid]>target){ return find(nums,target,start,mid-1); }else{ return find(nums,target,mid+1,end); } } }
[leedcode 35] Search Insert Position
原文：http://www.cnblogs.com/qiaomu/p/4633927.html