【leetcode】35搜索插入位置（自己理解的三種解法）

給定一個排序數組和一個目標值，在數組中找到目標值，並返回其索引。如果目標值不存在於數組中，返回它將會被按順序插入的位置。

你可以假設數組中無重複元素。

示例 1:

輸入: [1,3,5,6], 5
輸出: 2
示例 2:

輸入: [1,3,5,6], 2
輸出: 1
示例 3:

輸入: [1,3,5,6], 7
輸出: 4
示例 4:

輸入: [1,3,5,6], 0
輸出: 0

利用二分查找來做，順便貼出自己的二分查找代碼

package bite04;

/**
 * ⊙﹏⊙&&&&&&⊙▽⊙
 *舉例：區間長度[2,2] = 1，[2,2) = 0
 * 帶等於號：left+1 = right（結束條件）
 * 不帶等於號    left = right（結束條件）
 * @Auther: pangchenbo
 * @Date: 2020/1/7 09:52
 * @Description:二分查找
 */
public class binaryTest {
    /**
     * [left,right)搜索一次之後的區間爲[left,mid)和[mid+1,right)
     * @param arr
     * @param value
     * @return
     */
    public static int binarySerach(int[] arr,int value){
        //[left,right)
        int left = 0;
        int right = arr.length;
        while (left<right){//表示現在的左閉右開區間還有數
            int mid = (left+right)/2;
            if(arr[mid] == value)
                return mid;
            if(arr[mid]<value)
                left = mid+1;
            else
                right = mid;
        }
        return -1;
    }
    /**
     *搜索一次區間[left,mid-1]   [mid+1,right]
     * @param arr
     * @param value
     * @return
     */
    public static int binarySerach2(int[] arr,int value){
        //[left,right]
        int left = 0;
        int right = arr.length-1;//控制區間形式
        while (left<=right){//表示現在的左閉右閉區間還有數
            int mid = (left+right)/2;//int mid = left + (right - left) / 2;
            System.out.println("mid: "+mid);
            if(arr[mid] == value)
                return mid;
            if(arr[mid]<value) {
                left = mid + 1;
                System.out.println("left: "+left);
            }
            else {
                right = mid - 1;
                System.out.println("right: "+right);
            }
        }
        return -1;
    }

    public static void main(String[] args) {
        int [] array = {1,3,5,6};
        binarySerach2(array,2);
    }
}

本題代碼

解法一：因爲代碼的結束條件是left == right，所以返回right和left都可以

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

解法二：本題的結束條件是left+1 = right，在跑到[left,right]的時候還有數，然後進行left+1操作，所以最後返回right+1.

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

解法三：解法二的基礎上正常返回left

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