find peak or drop

来源:互联网 发布:盐城金石计价软件 编辑:程序博客网 时间:2024/05/16 11:33

//http://www.1point3acres.com/bbs/forum.php?mod=viewthread&tid=156077&extra=page%3D2%26filter%3Dsortid%26sortid%3D311%26searchoption%5B3046%5D%5Bvalue%5D%3D2%26searchoption%5B3046%5D%5Btype%5D%3Dradio%26sortid%3D311
找打peak or drop,原来用的是递归,这次改为迭代

public static int helper1(int[] nums) {    if (nums.length == 0)            return -1;        int len = nums.length;        if (Math.abs(nums[len - 1] - nums[0]) == len - 1)            return -1;// There is no such a valley or peak        int start = 0, end = nums.length - 1;        while (start < end) {            int mid = start + (end - start) / 2;            if ((nums[mid - 1] - nums[mid]) * (nums[mid + 1] - nums[mid]) > 0)                return nums[mid];            int diffOfIndex = mid - start;            int diffOfValue = Math.abs(nums[mid] - nums[start]);            //因为如果不相差,则下标和值是对应的,则没有发生波峰或波谷            if (diffOfIndex == diffOfValue)                start = mid;            else                end = mid;        }        return -1;}public static void main(String[] args) {int[] nums1 = { 1, 2, 3, 4, 3, 2 };int[] nums2 = {};int[] nums3 = { 1, 2, 3, 4, 5, 6, 7, 8, 7 };int[] nums4 = { 1, 2, 3, 4, 5 };int[] nums5 = { 5, 4, 3, 2, 1, 2, 3 };int[] nums6 = { 9, 8, 7, 6, 7 };System.out.println(find(nums1));System.out.println(find(nums2));System.out.println(find(nums3));System.out.println(find(nums4));System.out.println(find(nums5));System.out.println(find(nums6));System.out.println(helper1(nums1));        System.out.println(helper1(nums2));        System.out.println(helper1(nums3));        System.out.println(helper1(nums4));        System.out.println(helper1(nums5));        System.out.println(helper1(nums6));}


0 0
原创粉丝点击