Search for a Range 有序数组里查找一个数的出现区间 @LeetCode

最新推荐文章于 2025-02-14 14:54:14 发布

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最新推荐文章于 2025-02-14 14:54:14 发布

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该博客介绍了如何使用二分查找技术，在有序数组中高效地寻找目标数的起始和结束位置，详细阐述了经典的多次二分法解题思路。

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经典多次二分法！

package Level4;

import java.util.Arrays;

/**
 * Search for a Range
 * 
 * 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].
 * 
 */
public class S34 {

	public static void main(String[] args) {
		int[] A = {5,8,8,8,8,8,8,8,8,10};
		int[] ret = searchRange(A, 10);
		System.out.println(Arrays.toString(ret));
	}

	public static int[] searchRange(int[] A, int target) {
		int[] ret = {Integer.MAX_VALUE, Integer.MIN_VALUE};
		rec(A, target, ret, 0, A.length-1);
		if(ret[0] == Integer.MAX_VALUE){
			ret[0] = -1;
		}
		if(ret[1] == Integer.MIN_VALUE){
			ret[1] = -1;
		}
		return ret;
	}
	
	// 先用二分法找到满足条件，然后对两边分别二分法继续寻找
	public static void rec(int[] A, int target, int[] ret, int low, int high){
		if(low > high){
			return;
		}
		
		int mid = low + (high-low)/2;
		if(target == A[mid]){
			ret[0] = Math.min(ret[0], mid);		// 保存最小下限
			ret[1] = Math.max(ret[1], mid);		// 保存最大上限
			rec(A, target, ret, low, mid-1);		// 继续找满足条件的下限
			rec(A, target, ret, mid+1, high);		// 继续找满足条件的上限
		}else if(target < A[mid]){
			rec(A, target, ret, low, mid-1);
		}else{
			rec(A, target, ret, mid+1, high);
		}
	}

}

public class Solution {
    public int[] searchRange(int[] A, int target) {
        int[] ret = {-1,-1};
        rec(A, target, ret, 0, A.length-1);
        return ret;
    }
    
    public void rec(int[] A, int target, int[] ret, int left, int right) {
        if(left > right) {
            return;
        }
        
        int mid = left + (right-left)/2;
        if(A[mid] == target) {
            if(mid-1>=0 && A[mid-1] == target) {
                rec(A, target, ret, left, mid-1);
            } else {
                ret[0] = mid;
            }
            if(mid+1<=A.length-1 && A[mid+1] == target) {
                rec(A, target, ret, mid+1, right);
            } else {
                ret[1] = mid;
            }
            if(ret[0] != -1 && ret[1] != -1){
                return;
            }
        } else if(A[mid] < target) {
            rec(A, target, ret, mid+1, right);
        } else {
            rec(A, target, ret, left, mid-1);
        }
    }
}