[LeetCode]: 35: Search Insert Position

题目：

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

思路1：数组从小到大直接判断

代码：

    public static int searchInsert(int[] nums, int target) {

        if(target < nums[0]){
            return 0;
        }
        
        if(target > nums[nums.length-1]){
            return nums.length;
        }
        
        for(int i = 0;i<nums.length;i++){
            if(target == nums[i]){
                return i;
            }
            else{
                if(i>0){
                    if(target > nums[i-1] &&  target < nums[i]){
                        return i;
                    }
                }
            }
        }
        
        return -1;
    }

思路2：二分查找

    public static int searchInsert(int[] nums, int target) {
        if(target < nums[0]){
            return 0;
        }
        
        if(target > nums[nums.length-1]){
            return nums.length;
        }
        
        int low = 0;
        int high = nums.length-1;
        int middle =0;
        
        while(low <=high ){
            middle = (int)((low + high)/2);
            
            if(target == nums[middle]){
                return middle;
            }
            if(target < nums[middle]){
                high = middle-1;    
            }
            
            if(target > nums[middle]){
                low = middle+1;    
            }
        }
        
        return low;
    }