leetcode 34. Search for a Range

最新推荐文章于 2024-03-22 00:51:13 发布

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最新推荐文章于 2024-03-22 00:51:13 发布

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本文介绍了一种在已排序数组中找到特定目标值起始和结束位置的有效算法。通过两次二分查找实现，确保了运行时间复杂度为O(log n)。提供了具体的Java实现代码示例。

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Given an array of integers 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].

For example,
Given [5, 7, 7, 8, 8, 10] and target value 8,
return [3, 4].

做两次二分查找，寻找左右的边界

不过自己最后有点绕进去了，因为两个相邻的数相加数/2总是偏小的那个数。。。 = =

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

可以在第二个mid求取时+1使其向右偏，可以省下面的判断

另一种求该数位置和该数+1位置的二分法思路

public class Solution {
	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;
	}
}