Jump Game I and II leetcode

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.

当前的A[i]值代表，当前能跳的最大值，并不是一定要跳这么远的，比如A[0]=3，那么我们可以跳0到3步。

顺序扫描数组，记录下能够到达的最远位置:

class Solution {public:    bool canJump(int A[], int n) {        if(A == nullptr || n<= 0)        return false;        int i = 0;        int sum = 0;        while(i<n-1 && i<= sum) {            if(i+A[i]>sum)            sum = i+A[i];            if(sum>=(n-1))            break;            i++;        }        return (sum >= (n-1));    }};

way2:动态规划，状态转移数组state[i]代表当前从当前节点可跳到的最大节点，如果state[i]<=i&&i!=0，返回false

Jump Game 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 is 2. (Jump 1 step from index 0 to 1, then 3 steps to the last index.)

扫描数组，以确定当前最远能覆盖的节点，放入curr。然后继续扫描，直到当前的路程超过了上一次算出的覆盖范围，那么更新覆盖范围，同时更新条数，因为我们是经过了多一跳才能继续前进的。:

class Solution {public:    int jump(int A[], int n) {        int res = 0;        int last = 0;        int cur = 0;        for(int i=0;i<n;i++) {            if(i>last) {                res++;                last = cur;            }            cur = cur>(i+A[i])? cur:(i+A[i]);        }        return res;    }};