Paste_Image.png

My code:

public class Solution {
    public int searchInsert(int[] nums, int target) {
        if (nums == null || nums.length == 0)
            return 0;
        int lo = 0;
        int hi = nums.length - 1;
        return BinarySearch(lo, hi, target, nums);
    }
    
    private int BinarySearch(int lo, int hi, int target, int[] nums) {
        if (lo > hi)
            return lo;
        int mid = (lo + hi) / 2;
        if (nums[mid] == target)
            return mid;
        else if (nums[mid] > target)
            return BinarySearch(lo, mid - 1, target, nums);
        else
            return BinarySearch(mid + 1, hi, target, nums);
    }
    
    public static void main(String[] args) {
        Solution test = new Solution();
        int[] a = {1, 3, 5, 6};
        System.out.println(test.searchInsert(a, 0));
        
    }
}

My test result:

Paste_Image.png

這次題目比較簡單， 主要就是一個二分查找的實現(xiàn)。
這也算是我第一次實現(xiàn) binary search. 主要一個細(xì)節(jié)，就是， lo > hi 的時候再返回lo。

**
總結(jié)： binary search
**

My code:

public class Solution {
    public int searchInsert(int[] nums, int target) {
        if (nums == null || nums.length == 0)
            return 0;
        int begin = 0;
        int end = nums.length - 1;
        while (begin <= end) {
            int middle = begin + (end - begin) / 2;
            if (nums[middle] < target)
                begin = middle + 1;
            else if (nums[middle] > target)
                end = middle - 1;
            else
                return middle;
        }
        return begin;
    }
}

這道題木還是比較簡單的。

Anyway, Good luck, Richardo!

My code:

public class Solution {
    public int searchInsert(int[] nums, int target) {
        if (nums == null || nums.length == 0) {
            return -1;
        }
        
        int begin = 0;
        int end = nums.length - 1;
        
        while (begin <= end) {
            int mid = begin + (end - begin) / 2;
            if (nums[mid] > target) {
                end = mid - 1;
            }
            else if (nums[mid] < target) {
                begin = mid + 1;
            }
            else {
                return mid;
            }
        }
        
        return begin;
    }
}

簡單題。

Anyway, Good luck, Richardo! -- 09/01/2016