[LeetCode] 45. Jump Game II 跳跃游戏 II

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.

Example:

Input: [2,3,1,1,4]
Output: 2
Explanation: 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.

Note:

You can assume that you can always reach the last index.

解法1:贪婪Greedy

解法2:BFS

Java:

class Solution {
    public int jump(int[] A) {
        int sc = 0;
        int e = 0;
        int max = 0;
        for(int i=0; i<A.length-1; i++) {
            max = Math.max(max, i+A[i]);
            if( i == e ) {
                sc++;
                e = max;
            } 
        }
        return sc;
    }
}  

Python: BFS

class Solution:
    # @param {integer[]} nums
    # @return {integer}
    def jump(self, nums):
        n, start, end, step = len(nums), 0, 0, 0
        while end < n - 1:
            step += 1
            maxend = end + 1
            for i in range(start, end + 1):
                if i + nums[i] >= n - 1:
                    return step
                maxend = max(maxend, i + nums[i])
            start, end = end + 1, maxend
        return step  

C++: BFS

class Solution {
public:
    int jump(int A[], int n) {
         if(n<2)return 0;
         int level=0,currentMax=0,i=0,nextMax=0;

         while(currentMax-i+1>0){		//nodes count of current level>0
             level++;
             for(;i<=currentMax;i++){	//traverse current level , and update the max reach of next level
                nextMax=max(nextMax,A[i]+i);
                if(nextMax>=n-1)return level;   // if last element is in level+1,  then the min jump=level 
             }
             currentMax=nextMax;
         }
         return 0;
     }
};

  

 

 

[LeetCode] 55. Jump Game 跳跃游戏 

All LeetCode Questions List 题目汇总

posted @ 2018-09-13 15:30  轻风舞动  阅读(403)  评论(0编辑  收藏  举报