55. Jump Game

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], return true.

A = [3,2,1,0,4], return false.

代码：

class Solution {
public:
    //递归解法
    bool canJump2(vector<int>& nums) {
        return helper(nums, nums.size() - 1);
    }
    bool helper(vector<int>& nums, int last){
        if(last == 0) return true;
        int leftmost = nums.size();
        for(int i = last - 1; i >= 0; i--){
            if(nums[i] >= last - i)
            {leftmost = i;
            break;//这个break太重要了！ 不用每次都遍历数组寻找到leftmost，只需要找到left即可，因为一定会有leftmost可以到达left
            }
        }
        if(leftmost == nums.size()) return false;
        return helper(nums, leftmost);
    }
    //迭代解法
    bool canJump(vector<int>& nums) {
        int reach = 0;
        for(int i = 0; i <= reach; i++){
            reach = max(i + nums[i], reach);
            if(reach >= nums.size() - 1)return true;
        }
        return false;
    }
};

最优化代码：

bool canJump(int A[], int n) {
    int i = 0;
    for (int reach = 0; i < n && i <= reach; ++i)
        reach = max(i + A[i], reach);
    return i == n;
}

题目2：

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.)

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

代码：

class Solution {
public:
    //TLE
    int jump2(vector<int>& nums) {
        int times = 0;
        unordered_set<int> start;
        start.insert(0);
        
        if(nums.size() == 1) return 0;
        while(1){
             unordered_set<int> temp;
            times++;
            for(int i : start)
            
            {
                
                for(int j = 1; j <= nums[i]; j++){
                    if(!temp.count(i + j))
                    temp.insert(i + j);
                }
            }
            start = temp;
            if(temp.count(nums.size() - 1)) break;
            
        }
        return times;
    }
    //AC code
    int jump(vector<int>& nums) {
        int bound = 0, reach = 0, level = 0;
        if(nums.size() == 1) return 0;
        for(int i = 0; i < nums.size() && i <= bound; i++){
            reach = max(reach, i + nums[i]);
            if(i == bound && i != nums.size() - 1){
                bound = reach;
                level++;
            }
        }
        return level;
    }
};