Total Occurrence of Target

最新推荐文章于 2025-05-12 15:33:14 发布

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最新推荐文章于 2025-05-12 15:33:14 发布

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分类专栏： Binary Search

本文链接：https://blog.youkuaiyun.com/u013325815/article/details/103664060

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本文介绍了一种在升序整数数组中查找特定目标数总出现次数的方法，通过二分查找找到目标数的最后一个位置，然后逆向统计相同元素的数量，实现了O(logn)的时间复杂度。

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Given a target number and an integer array sorted in ascending order. Find the total number of occurrences of target in the array.

Example

Example1:

Input: [1, 3, 3, 4, 5] and target = 3, 
Output: 2.

Example2:

Input: [2, 2, 3, 4, 6] and target = 4, 
Output: 1.

Example3:

Input: [1, 2, 3, 4, 5] and target = 6, 
Output: 0.

Challenge

Time complexity in O(logn)

思路：先find到target index，如果-1直接return 0.否则再统计；

public class Solution {
    /**
     * @param A: A an integer array sorted in ascending order
     * @param target: An integer
     * @return: An integer
     */
    public int totalOccurrence(int[] A, int target) {
        if(A == null || A.length == 0) {
            return 0;
        }
        
        int i = findLastIndex(A, target);
        if(i == -1) {
            return 0;
        } 
        int count = 0;
        while(i >= 0) {
            if(A[i] == target){
                count++;
                i--;
            } else {
                break;
            }
        }
        return count;
    }
    
    private int findLastIndex(int[] A, int target) {
        int start = 0; int end = A.length - 1;
        while(start + 1 < end) {
            int mid = start + (end - start) / 2;
            if(A[mid] <= target) {
                start = mid;
            } else {
                end = mid;
            }
        }
        if(A[end] == target) {
            return end;
        } 
        if(A[start] == target){
            return start;
        }
        return -1;
    }
}