45. Jump Game II
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Mean:
给定一个数组a,玩家的初始位置在idx=0,玩家需要到达的位置是idx=a.size()-1.
如果玩家在idx处,那么他最多可以向后走a[idx]步,问最少多少步可以走到终点.
analyse:
方法非常巧妙,类似于BFS.
Time complexity: O(N)
view code
/** * ----------------------------------------------------------------- * Copyright (c) 2016 crazyacking.All rights steperved. * ----------------------------------------------------------------- * Author: crazyacking * Date : 2016-03-08-14.52 */ #include <queue> #include <cstdio> #include <set> #include <string> #include <stack> #include <cmath> #include <climits> #include <map> #include <cstdlib> #include <iostream> #include <vector> #include <algorithm> #include <cstring> using namespace std; typedef long long( LL); typedef unsigned long long( ULL); const double eps( 1e-8); class Solution { public : int jump( vector < int >& nums) { int i( 0 ), j( 1 ), steps( 0 ), N( nums . size()); while( j < N) { int endj = min( nums [ i ] + i , N); while( j < = endj) { if( nums [ j ] + j > nums [ i ] + i) i = j; j ++; } steps ++; } return steps; } }; // this is my TLE code :( //class Solution //{ //public: // int jump(vector<int>& nums) // { // queue<pair<int,int>> que; // pos,step // int res=INT_MAX; // que.push(make_pair(0,0)); // while(!que.empty()) // { // pair<int,int> top=que.front(); // que.pop(); // if(top.first>=nums.size()-1) // res=min(res,top.second); // for(int i=1;i<=nums[top.first] && i+top.first<=nums.size();++i) // { // que.push(make_pair(top.first+i,top.second+1)); // } // } // return res; // } //}; int main() { int n; while( cin >>n) { vector < int > ve; for( int i = 0; i <n; ++ i) { int tempVal; cin >> tempVal; ve . push_back( tempVal); } Solution solution; int ans = solution . jump( ve); cout << ans << endl; } return 0; } /* */