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/*/*******************************************************************************
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** **
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** Jiedi(China nanjing)Ltd. **
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** 创建:丁宋涛 夏曹俊,此代码可用作为学习参考 **
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*******************************************************************************/
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/*****************************FILE INFOMATION***********************************
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**
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** Project : 算法设计与编程实践---基于leetcode的企业真题库
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** Contact : xiacaojun@qq.com
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** 博客 : http://blog.csdn.net/jiedichina
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** 视频课程 : 网易云课堂 http://study.163.com/u/xiacaojun
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腾讯课堂 https://jiedi.ke.qq.com/
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csdn学院 https://edu.csdn.net/course/detail/25037
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** 51cto学院 http://edu.51cto.com/lecturer/index/user_id-100013755.html
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** 老夏课堂 http://www.laoxiaketang.com
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**
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** 算法设计与编程实践---基于leetcode的企业真题库 课程群 :296249312 加入群下载代码和交流
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** 微信公众号 : jiedi2007
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** 头条号 : 夏曹俊
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**
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*****************************************************************************
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//!!!!!!!!! 算法设计与编程实践---基于leetcode的企业真题库 课程 QQ群:296249312 下载代码和交流*/
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/*
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* @lc app=leetcode.cn id=697 lang=cpp
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*
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* [697] 数组的度
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*/
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#include <vector>
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#include <map>
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using namespace std;
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class Solution {
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public:
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int findShortestSubArray(vector<int>& nums) {
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int n = nums.size();
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map<int,vector<int>> tmp;//vector记录出现的相同数的位置
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for(int i=0;i<n;i++){
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tmp[nums[i]].push_back(i);
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}
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int count = 0; //表征数组的度
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int res = INT_MAX;
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for(auto iter=tmp.begin();iter!=tmp.end();iter++){
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int m = iter->second.size();//算一下数组的度
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if(m>count){
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count = m;
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res = iter->second[m-1]-iter->second[0]+1;
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}
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if(m==count){
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res=min(res,iter->second[m-1]-iter->second[0]+1);
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}
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}
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return res;
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}
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};
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