二分查找算法

二分查找算法的具体思想我就不再阐述了，要进行二分查找的前提是数组有序

public class Solution {

    public static void main(String[] args) {
        Solution solution = new Solution();
        int[] array = {1, 2, 5, 7};
        int num = solution.binarySearch(array, 5);
        // 2
        System.out.println(num);
        num = solution.binarySearch(array, 10);
        // -1
        System.out.println(num);
    }

    public int binarySearch(int[] array,int search) {
        int start = 0;
        int end = array.length - 1;
        while (start <= end) {
            int mid = (start + end) >> 1;
            if (array[mid] == search) {
                return mid;
            } else if (array[mid] < search) {
                start = mid + 1;
            } else {
                end = mid - 1;
            }
        }
        return -1;
    }
}

二分查找算法的变种

查找第一个值等于给定值的情况

二分查找算法能返回数组中给定元素的下标，但不能保证是第一个，因为数组中的元素有可能重复，所以找到指定元素后可以继续二分

public class Solution {

    public static void main(String[] args) {
        Solution solution = new Solution();
        int[] array = {1, 1, 1, 3, 5};
        int num = solution.binarySearch(array, 5);
        // 4
        System.out.println(num);
        num = solution.binarySearch(array, 1);
        // 0
        System.out.println(num);
        num = solution.binarySearch(array, 2);
        // -1
        System.out.println(num);
    }

    public int binarySearch(int[] array,int search) {
        int start = 0;
        int end = array.length - 1;
        while (start <= end) {
            int mid = (start + end) >> 1;
            if (array[mid] == search) {
                if (mid == start || array[mid - 1] != search) {
                    return mid;
                } else {
                    end = mid - 1;
                }
            } else if (array[mid] < search) {
                start = mid + 1;
            } else {
                end = mid - 1;
            }
        }
        return -1;
    }
}

查找最后一个值等于给定值的情况

和上面类似

public class Solution {

    public static void main(String[] args) {
        Solution solution = new Solution();
        int[] array = {1, 1, 1, 3, 5};
        int num = solution.binarySearch(array, 5);
        // 4
        System.out.println(num);
        num = solution.binarySearch(array, 1);
        // 2
        System.out.println(num);
        num = solution.binarySearch(array, 2);
        // -1
        System.out.println(num);
    }

    public int binarySearch(int[] array,int search) {
        int start = 0;
        int end = array.length - 1;
        while (start <= end) {
            int mid = (start + end) >> 1;
            if (array[mid] == search) {
                if (mid == end || array[mid + 1] != search) {
                    return mid;
                } else {
                    start = mid + 1;
                }
            } else if (array[mid] < search) {
                start = mid + 1;
            } else {
                end = mid - 1;
            }
        }
        return -1;
    }
}

查找第一个大于等于给定值的情况

查找最后一个小于等于给定值的情况

参考博客

[1]https://mp.weixin.qq.com/s/YbAfGfGdy0WEdEWKV0B5Cg