leetcode 35. Search Insert Position

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选择匿名的用户   2021-6-2 20:51   1249   0

一 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.

Example 1:

Input: [1,3,5,6], 5
Output: 2

Example 2:

Input: [1,3,5,6], 2
Output: 1

Example 3:

Input: [1,3,5,6], 7
Output: 4

Example 4:

Input: [1,3,5,6], 0
Output: 0

二 分析

EASY级别,题目要求给定一个target,和有序序列,如果target在序列中,则返回其索引,否则给出当插入target且不改变序列性质时索引位置。

循环法

因为已经排序的数组,只需target与数组当前位比较,如果小于等于当前位,则返回当前位的下标。一开始做了边界判断。

class Solution {
    public int searchInsert(int[] nums, int target) {
        int res =0;
  if(nums== null||nums.length==0 ||target<= nums[0] ){
    return res;
   }

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

应该是LeetCode抽了。这是O(N)的时间复杂度。

Runtime: 0 ms, faster than 100.00% of Java online submissions for Search Insert Position.

Memory Usage: 38.1 MB, less than 100.00% of Java online submissions for Search Insert Position.

二分查找

二分应该这是题目要考察的点吧,上面的方法应该不是真正的目的。

public static void main(String[] args) {
  // TODO Auto-generated method stub
  int[]  nums6 ={2,7,8,9,10};
  System.out.println(searchInsert(nums6,9));
  
  int[]  nums4 ={1,3,5};
  System.out.println(searchInsert(nums4,4));
  int[]  nums5 ={1,3,5};
  System.out.println(searchInsert(nums5,2));
  int[]  nums1 ={1,3,5,6};
  System.out.println(searchInsert(nums1,2));
 
  int[]  nums ={1,3,5,6};
  System.out.println(searchInsert(nums,5));
 
  int[]  nums2 ={1,3,5,6};
  System.out.println(searchInsert(nums2,7));
  int[]  nums3 ={1,3,5,6};
  System.out.println(searchInsert(nums3,0));
 }

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

Runtime: 0 ms, faster than 100.00% of Java online submissions for Search Insert Position.

Memory Usage: 38 MB, less than 100.00% of Java online submissions forSearch Insert Position.

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