算法第四版1.4.10 修改二分查找算法，使之总是返回和被查找的键匹配的索引最小的元素

题目

1.4.10 修改二分查找算法，使之总是返回和被查找的键匹配的索引最小的元素。（且仍能够保证对数级别的运行时间）

1.4.10 Modify binary search so that it always returns the element with the smallest index that matches the search element (and still guarantees logarithmic running time).

代码

package homework1_4;

/**
 * @description: ${description}
 * @create: 2019-03-05
 **/
public class W_1_4_10 {
    //数组a必须是有序的,原始二分查找法
    public static int rank(int key, int[] a) {
        int low = 0;
        int high = a.length - 1;
        while (low <= high) {
            int mid = low + (high - low) / 2;
            if (a[mid] < key) {
                low = mid + 1;
            } else if (a[mid] > key) {
                high = mid - 1;
            } else {
                return mid;
            }
        }
        return -1;
    }

    //1.4.10的做法1 ,一般情况可以满足对数
    public static int rank1(int key, int[] a) {
        int low = 0;
        int high = a.length - 1;
        while (low <= high) {
            int mid = low + (high - low) / 2;
            if (a[mid] < key) {
                low = mid + 1;
            } else if (a[mid] > key) {
                high = mid - 1;
            } else {
                while (a[mid] == a[mid - 1]) {
                    mid--;
                    if (mid < 0) {
                        break;
                    }
                }
                return mid;
            }
        }
        return -1;
    }

    //1.4.10的做法2 高级改进版本
    public static int rank2(int key, int[] a) {
        int low = 0;
        int high = a.length - 1;
        while (low <= high) {
            int mid = low + (high - low) / 2;
            if (a[mid] < key) {
                low = mid + 1;
            } else if (a[mid] > key) {
                high = mid - 1;
            } else if (a[mid - 1] == a[mid] && mid > 0) {
                high = mid;
            } else {
                return mid;
            }
        }
        return -1;
    }

    public static void main(String[] args) {
        int key = 5;
        int[] a = {1, 2, 3, 4, 5, 5, 5, 5, 5, 5, 5, 6, 7, 8, 9, 10, 11};
        System.out.println(rank(key, a));
        System.out.println(rank1(key, a));
        System.out.println(rank2(key, a));
    }
}

结果