Search for a Range

最新推荐文章于 2022-03-08 23:18:06 发布

原创最新推荐文章于 2022-03-08 23:18:06 发布 · 585 阅读

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本文介绍了一种算法，能够在已排序的整数数组中找到给定目标值的起始和结束位置，符合O(log n)的时间复杂度要求。如果未找到目标值，则返回[-1,-1]。例如，在数组[5,7,7,8,8,10]中查找8，将返回[3,4]。

Given a sorted array of integers, 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]

Code:

public class Solution {
    public int[] searchRange(int[] A, int target) {
        int i = 0;
        int j = A.length - 1;
        int index = getIndex(A, target);
        int[] indexes = new int[2];
        if (index == -1) {
            indexes[0] = -1;
            indexes[1] = -1;
            return indexes;
        } else {
            indexes[0] = getLowerIndex(A, target, index);
            indexes[1] = getUpperIndex(A, target, index);
        }
        return indexes;
        
    }
    
    public int getUpperIndex(int[] A, int target, int startPosition) {
        int start = A.length - 1;
        while (!(startPosition == A.length - 1 || A[startPosition + 1] != target)) {
            int mid = (int) ((start + startPosition) / 2);
            if (A[mid] == target) {
                if (startPosition == mid) {
                    startPosition = mid + 1;
                } else {
                    startPosition = mid;
                }
                
            } else {
                start = mid - 1;
            }
        }
        return startPosition;
    }
    
    public int getLowerIndex(int[] A, int target, int startPosition) {
        int start = 0;
        while (!(startPosition == 0 || A[startPosition - 1] != target)) {
            int mid = (int) ((start + startPosition) / 2);
            if (A[mid] == target) {
                startPosition = mid;
            } else {
                start = mid + 1;
            }
        }
        return startPosition;
    }
    
    public int getIndex(int[] A, int target) {
        int i = 0;
        int j = A.length - 1;
        while (i <= j) {
            int mid = (int) ((i + j) / 2);
            if (A[mid] == target) {
                return mid;
            } else if (A[mid] > target) {
                j = mid - 1;
            } else {
                i = mid + 1;
            }
        } 
        return -1;
    }
}