Leetcode 34. Find First and Last Position of Element in Sorted Array

最新推荐文章于 2024-11-08 13:05:39 发布

周英俊520

最新推荐文章于 2024-11-08 13:05:39 发布

阅读量146

点赞数

CC 4.0 BY-SA版权

分类专栏： leetcode

本文链接：https://blog.youkuaiyun.com/qq_38835878/article/details/93176286

leetcode 专栏收录该内容

18 篇文章

订阅专栏

博客围绕在升序整数数组中查找给定值的起始和结束位置展开，要求算法时间复杂度为O(log n)。介绍了两种解法，一是写demo找最小索引并调用两次，二是变种二分查找，还提及二分查找中while循环条件的编程细节。

摘要生成于 C知道，由 DeepSeek-R1 满血版支持，前往体验 >

Given an array of integers nums sorted in ascending order, find the starting and ending position of a given target value.

Your algorithm's runtime complexity must be in the order of O(log n).

If the target is not found in the array, return [-1, -1].

Example 1:

Input: nums = [5,7,7,8,8,10], target = 8
Output: [3,4]

Example 2:

Input: nums = [5,7,7,8,8,10], target = 6
Output: [-1,-1]

这道题有两种解法，第一是写一个demo, 这个demo能找到target在数组当中的最小索引，然后调用两次demo，传入的参数分别是target 和 target + 1. 即可解决。第二种解法是变种的二分查找. (另外在写二分查找的时候while循环中lo < hi 还是lo <=hi.取决于hi的定义) 我的习惯hi = nums.length - 1这时候就是lo <= hi, 因为不这样的话当数组只有1个元素的时候他就不会进入while循环，这个元素就没有被查找。一般的二分查找这样是没问题的，但是在这道题当中查不到元素要返回的是-1，所以你必须进入这个while循环，这也是一些编程细节的问题。

// 方法1
public int[] searchRange(int[] A, int target) {
		int start = Solution.firstGreaterEqual(A, target);
		if (start == A.length || A[start] != target) {
			return new int[]{-1, -1};
		}
		return new int[]{start, Solution.firstGreaterEqual(A, target + 1) - 1};
	}

	//find the first number that is greater than or equal to target.
	//could return A.length if target is greater than A[A.length-1].
	//actually this is the same as lower_bound in C++ STL.
	private static int firstGreaterEqual(int[] A, int target) {
		int low = 0, high = A.length;
		while (low < high) {
			int mid = low + ((high - low) >> 1);
			//low <= mid < high
			if (A[mid] < target) {
				low = mid + 1;
			} else {
				//should not be mid-1 when A[mid]==target.
				//could be mid even if A[mid]>target because mid<high.
				high = mid;
			}
		}
		return low;
	}

// 方法2
public int[] searchRange(int[] nums, int target) {
        int lo = 0; 
        int[] res = new int[2];
        int left = findFirstK(nums, target);
        int right = findLastK(nums, target);
        res[0] = left;
        res[1] = right;
        return res;
    }
    
    public int findFirstK(int[] nums, int target) {
        int lo = 0;
        int hi = nums.length - 1;
        while( lo <= hi) {
            int mid = (lo + hi) / 2;
            if(nums[mid] < target){
                lo = mid + 1;
            } else if(nums[mid ] > target) {
                hi = mid - 1;
            } else if(mid - 1 >= 0 && nums[mid - 1] == target) {
                hi = mid - 1;
            } else {
                return mid;
            }
        }
        return -1;
    }
    
    public int findLastK(int[] nums, int target) {
        int lo = 0;
        int hi = nums.length - 1;
        while( lo <= hi) {
            int mid = (lo + hi) / 2;
            if(nums[mid] < target){
                lo = mid + 1;
            } else if(nums[mid ] > target) {
                hi = mid - 1;
            } else if(mid + 1 < nums.length && nums[mid + 1] == target) {
                lo = mid + 1;
            } else {
                return mid;
            }
        }
        return -1;
    }