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[LeetCode] Jump Game II(贪婪算法)
Given an array of non-negative integers, you are initially positioned at the first index of the array.
Each element in the array represents your maximum jump length at that position.
Your goal is to reach the last index in the minimum number of jumps.
For example: Given array A = [2,3,1,1,4]
The minimum number of jumps to reach the last index is 2. (Jump 1 step from index 0 to 1, then 3 steps to the last index.)
用queue实现bfs的思想，Memory Limited Exceeded！
然后把queue改成vector实现同样的思想，Time Limited Exceeded！
class Solution {public: int jump(int A[], int n) { if(n<2) return 0; pair<int,int> temp;//pair记录下标和到此下标的步数 vector<pair<int,int> > v1; int minStep = n ; temp = make_pair(0,0); v1.push_back(temp); while(!v1.empty()){ temp = v1.back(); int Index = temp.first; int step = temp.second; v1.pop_back(); if(Index>=n-1){ if(step == 2) return 2; else if(step<minStep) minStep = step; continue; } int newIndex = Index + A[Index]; if(newIndex == Index) continue; temp = make_pair(newIndex,++step); v1.push_back(temp); for(int i = Index+1;i<newIndex;i++){ if(i+A[i]<newIndex) continue; else{ temp = make_pair(i+A[i],step+1); v1.push_back(temp); } } }//end while return minStep; }//end func};
克服浪费时间和浪费空间的问题，Here is a solution from other，怎么会如此简洁（贪婪算法）：
/* * We use "last" to keep track of the maximum distance that has been reached * by using the minimum steps "ret", whereas "curr" is the maximum distance * that can be reached by using "ret+1" steps. Thus, * curr = max(i+A[i]) where 0 <= i <= last. */class Solution { //last和curr都是下标public: int jump(int A[], int n) { int ret = 0; int last = 0; int curr = 0; for (int i = 0; i < n; ++i) { if (i > last) { last = curr; ++ret; } curr = max(curr, i+A[i]); //经过ret+1步跳到的下标curr位置 } return ret; }};
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