在排序数组中查找数字（Java实现）

最新推荐文章于 2022-02-07 23:10:45 发布

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最新推荐文章于 2022-02-07 23:10:45 发布

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本文深入探讨了在排序数组中进行高效搜索的三种高级技巧：查找数字出现次数、寻找缺失数字以及查找数值等于下标的元素。通过具体的算法实现和示例，读者可以了解如何在实际应用中使用这些技巧。

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public class E53FindInNumbersInOrder {
    //在排序数组中查找数字

    /*问题一：数字在数组中出现的次数*/
    public static int getTimes(int[] numbers, int length, int target) {
        int times = -1;
        if (numbers != null && length > 0) {
            int start = getStartIndex(numbers, length, target);
            int end = getEndIndex(numbers, length, target);
            if (start > -1 && end > -1)
                times = end - start + 1;
        }
        return times;
    }

    //获取最后出现的索引
    private static int getEndIndex(int[] numbers, int length, int target) {
        int start = 0;
        int end = length - 1;
        while (start <= end) {
            int middle = (start + end) >> 1;
            if (numbers[middle] == target) {
                if (middle == length - 1 || (middle < length - 1 && numbers[middle + 1] != target))
                    return middle;
                else
                    start = middle + 1;
            } else if (numbers[middle] < target)
                start = middle + 1;
            else
                end = middle - 1;
        }
        return -1;
    }

    //获取首次出现的索引
    private static int getStartIndex(int[] numbers, int length, int target) {
        int start = 0;
        int end = length - 1;
        while (start <= end) {
            int middle = (start + end) >> 1;
            if (numbers[middle] == target) {
                if (middle == 0 || (numbers[middle - 1] != target))
                    return middle;
                else
                    end = middle - 1;
            }
            else if (numbers[middle] > target)
                end = middle - 1;
            else
                start = middle + 1;
        }
        return -1;
    }

    /*问题二：寻找排序数组中唯一缺失的数字*/
    public static int getLostNumber(int[] numbers, int length){
        if (numbers == null || length <= 0)
            return -1;
        int start = 0;
        int end = length - 1;
        while(start <= end){
            int middle = (start + end) >> 1;
            if (numbers[middle] != middle){
                if (middle == 0 || numbers[middle - 1] == middle - 1)
                    return middle;
                else
                    end = middle - 1;
            }
            else
                start = middle + 1;
        }
        return -1;
    }

    /*问题三：数组中数值等于下标的数字*/
    public static int getNumberEqualToIndex(int[] numbers, int length){
        if (numbers == null || length <= 0)
            return -1;
        int start = 0;
        int end = length - 1;
        while(start <= end){
            int middle = (start + end) >> 1;
            if (numbers[middle] == middle)
                return middle;
            else if (numbers[middle] > middle)
                end = middle - 1;
            else
                start = middle + 1;
        }
        return -1;
    }

    //测试用例
    public static void main(String[] args){
        /*数字出现的次数*/
        int[] numbers1 = {1, 2, 3, 3, 3, 4, 5};
        System.out.println(E53FindInNumbersInOrder.getTimes(numbers1, 7, 3)); //3
        System.out.println(E53FindInNumbersInOrder.getTimes(numbers1, 7, 1)); //3
        System.out.println(E53FindInNumbersInOrder.getTimes(numbers1, 7, 5)); //5
        /*缺失的数字*/
        int[] numbers2 = {0, 1, 2, 3, 4, 5, 7, 8 , 9};
        System.out.println(E53FindInNumbersInOrder.getLostNumber(numbers2, 9)); //6
        /*数组索引与数值相同的数字*/
        int[] numbers3 = {-3, -1, 1, 3, 5};
        System.out.println(E53FindInNumbersInOrder.getNumberEqualToIndex(numbers3, 5)); //3
    }
}