(Java)LeetCode-35. Search Insert Position

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

连续几道题都会用到二分查找了，这道题也不例外。只需要把前面几题用过的二分查找函数做稍许修改即可。当查找到最后一步时，若没有找到，不返回-1，要是小于target，则返回index+1，若大于target，则返回target，此题简单~代码如下：

public class Solution {
public int searchInsert(int[] nums, int target) {
return binarySearch(nums,target,0,nums.length-1);
}
private int binarySearch(int[] nums, int target, int left, int right) {
// TODO Auto-generated method stub
if(left > right)
return -1;
if(left == right){
if(nums[left] >= target){
return left;
}else{
return right+1;
}
}
int mid = (left + right)/2;
if(target < nums[mid]){
return binarySearch(nums,target,left,mid);
}else if(target == nums[mid]){
return mid;
}else{
return binarySearch(nums,target,mid+1,right);
}
}
}