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
只需要把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 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; } } //避免溢出 int middle = begin + (end - begin) / 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); } } public int searchInsert(int[] nums, int target) { int low = 0; int high = nums.length - 1; int middle = 0; while (low <= high) { middle = low + (high - low) / 2;//避免溢出 if (nums[middle] == target) return middle; else if (nums[middle] > target) high = middle - 1; else low = middle + 1; } return low; }