jump-game i&&ii 能否跳出区间 贪心

I:

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.

Determine if you are able to reach the last index.

For example:
A =[2,3,1,1,4], returntrue.

A =[3,2,1,0,4], returnfalse.

维护一个当前能跳到的最大值maxJump, 若是maxJump 已经>=nums.length-1, 说明能跳到最后一个点，return true.若是过程中maxJump <= i, 说明跳到当前点便不能往前，跳出loop, return false.

class Solution {
public:
    bool canJump(int A[], int n) {
        int cur=;
        for(int i=;i<n;++i)
        {
            if(i>cur ||cur>=n-)
                break;
            cur=max(cur,i+A[i]);
        }
        return cur>=n-;
    }
};

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.

For example:
Given array A =[2,3,1,1,4]

The minimum number of jumps to reach the last index is2. (Jump1step from index 0 to 1, then3steps to the last index.)

class Solution {
public:
    int jump(int A[], int n) {
        vector<int> dp(n,);// dp存放都到各点的最小步数

        for(int i=;i<n;++i)
        {
            int max=min(i+A[i],n-);// 从i点出发能走的最远距离
            for(int j=i+;j<=max;++j)
            {
                if(dp[j]==)
                    dp[j]=dp[i]+;// 如果位置没被走过,则到达j点的步数为dp[i]+1
            }
            if(dp[n-]!=) break;// 当第一次到达终点时,肯定是到达终点最短的步数
        }
        return dp[n-];
    }
};