[LeetCode]: 35: Search Insert Position

標籤：

題目：

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：數組從小到大直接判斷

 

代碼：

    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2：二分尋找

    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;    }

 

[LeetCode]: 35: Search Insert Position