描述

给 n 个整数的山脉数组，即先增后减的序列，找到山顶（最大值）

样例

给出数组为 [1, 2, 4, 8, 6, 3],返回 8
给出数组为 [10, 9, 8, 7]，返回 10

代码

方法1

public class Solution {
    /**
     * @param nums a mountain sequence which increase firstly and then decrease
     * @return then mountain top
     */
    public int mountainSequence(int[] nums) {
        if (nums == null || nums.length == 0) {
            return -1;
        }

        int start = 0;
        int end = nums.length - 1;
        while (start + 1 < end) {
            int mid = start + (end - start) / 2;
            /* 应注意条件语句写法
             *  若写成nums[mid] < nums[mid - 1] 和nums[mid] < nums[mid + 1];
             *  在mid是最大值时，start和end都不更新，死循环
             *  写下面if条件时要保证同时只满足一个条件
             *  不会出现同时满足两个和不满足两个的情形
             */
            // 保证mid取顶点时 ，不会出现死循环               
            if (nums[mid] > nums[mid - 1]) {
                start = mid;
            }
            if (nums[mid] > nums[mid + 1]) {
                end = mid;
            }
            // 不存在相等的点;否则二分法不能做
        }
        return Math.max(nums[start], nums[end]);
    }
}

方法2

// version 2: 一个比较啰嗦的版本，本质也是2分法，每次取两个点
public class Solution {
    /**
     * @param nums a mountain sequence which increase firstly and then decrease
     * @return then mountain top
     */
    public int mountainSequence(int[] nums) {
        int left = 0, right = nums.length - 1;
        while (left + 1 < right) {
            int m1 = left + (right - left) / 2;
            int m2 = right - (right - m1) / 2;
            if (nums[m1] < nums[m2]) {
                left = m1 + 1;
            } else if (nums[m1] > nums[m2]) {
                right = m2 - 1;
            } else {
                left = m1;
                right = m2;
            }
        }
        return nums[left] > nums[right] ? nums[left] : nums[right];
    }
}