【LintCode】Pattern（C语言实现）

题目描述

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Given a sequence of n integers a1, a2, …, an, a 132 pattern is a subsequence ai, aj, ak such that i < j < k and ai < ak < aj. Design an algorithm that takes a list of n numbers as input and checks whether there is a 132 pattern in the list.

n will be less than 20,000.

测试数据

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Given nums = [1, 2, 3, 4]
return False // There is no 132 pattern in the sequence.

Given nums = [3, 1, 4, 2]
return True // There is a 132 pattern in the sequence: [1, 4, 2].

解题思路

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找满足A[i] < A[k]

C实现代码

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#include <stdio.h>#define true 1#define false 0#define VSIZE 100#define SSIZE 100#define INT_MIN -10000000typedef unsigned char bool;typedef struct{    int data[VSIZE];    int length;}vector;typedef struct{    int data[SSIZE];    int top;}stack;bool solve(vector v){    int third = INT_MIN;    stack second;    int i;    second.top = -1;    for(i = v.length - 1; i >= 0; --i)    {        //当前的数比third小，也肯定比栈里的数小，条件满足        if(v.data[i] < third)            return true;        else        {            while(second.top != -1)            {                //如果当前的数比栈顶的数大，可以用栈顶的数更新                //third，从而得到一个更大的third，加快算法速度                if(second.data[second.top] < v.data[i])                {                    if(third < second.data[second.top])                        third = second.data[second.top];                    --second.top;                }                else                    break;            }            second.data[++second.top] = v.data[i];        }    }    return false;}int main(int argc, char *argv[]){    vector v;    int i;    while(~scanf("%d", &v.length))    {        for(i = 0; i < v.length; i++)            scanf("%d", &v.data[i]);        if(solve(v))            printf("YES\n");        else            printf("NO\n");    }    return 0;}