LintCode 1287 · Increasing Triplet Subsequence (贪心算法)

1287 · Increasing Triplet Subsequence
Algorithms
Medium

Description
Given an unsorted array return whether an increasing subsequence of length 3 exists or not in the array.

Formally the function should:
Return true if there exists i, j, k
such that arr[i] < arr[j] < arr[k] given 0 ≤ i < j < k ≤ n-1 else return false.
Your algorithm should run in O(n) time complexity and O(1) space complexity.

Example
Example1

Input: [1, 2, 3, 4, 5]
Output: true
Example2

Input: [5, 4, 3, 2, 1]
Output: false

解法1：参考的网上的答案。first和second分别表示当前所遇到的一小一大数的比较小的组合。如果碰到一个新数>second，就可以直接返回true。如果first<新数<=second，那么更新second，first不用更新。如果新数<=first，那么更新first=新数。

class Solution {
public:
    /**
     * @param nums: a list of integers
     * @return: return a boolean
     */
    bool increasingTriplet(vector<int> &nums) {
        int n = nums.size();
        if (n < 3) return false;
        int first = nums[0], second = INT_MAX;
        for (int i = 1; i < n; i++) {
            if (second < nums[i]) {
                return true;
            } else if (first < nums[i]) {
                second = nums[i];
            } else {
                first = nums[i];
            }
        }
        return false;
    }
};

也可以当first > nums[i]时，更新second=first, first=nums[i]。这样，first, second组合就是尽可能小的一小一大组合了。

class Solution {
public:
    /**
     * @param nums: a list of integers
     * @return: return a boolean
     */
    bool increasingTriplet(vector<int> &nums) {
        int n = nums.size();
        if (n < 3) return false;
        int first = nums[0], second = INT_MAX;
        for (int i = 1; i < n; i++) {
            if (second < nums[i]) {
                return true;
            } else if (first < nums[i]) {
                second = nums[i];
            } else if (first > nums[i]) {
                second = first;
                first = nums[i];
            }
        }
        return false;
    }
};