[C + +] leetcode:103 Jump Game (local optimal and global optimal method)

Title:

Given an array of non-negative integers, you is initially positioned at the first index of the array.

Each element of the array represents your maximum jump length is at that position.

Determine if you is able to reach the last index.

for example:
a = [2,3,1,1,4" , Return true .

A = [3,2,1,0,4] , return false .

idea: Dynamic programming solution, we maintain two variables, a local optimal, a global optimal. One represents the furthest distance we can reach so far, that is, local = A[i]+i, and the other is the furthest distance that can be jumped at the current step, global = max (Global, local). The recursive relationship is determined, the recursive termination condition reaches the end of the array, if global > n-1, we can use the end of the arrival array (global represents the maximum distance, we can choose exactly the step to reach the end).

Complexity: Traverse once, O (N), Space O (1)

Attention:

1. Empty array, return false

if (n = = 0) return false;

2. When jumping, we need to meet two conditions, I < n && i <= reach, we can not exceed the global maximum distance.

for (int i = 0; i < n && i <= reach; i++)

AC Code:

Class Solution {public:    bool Canjump (int a[], int n) {        if (n = = 0) return false;        int reach = 0; Global furthest distance                for (int i = 0; i < n && i <= reach; i++)        {            reach = max (a[i]+i, Reach);        }                if (Reach >= n-1)            return true;        else            return false;    }};

[c++]leetcode:103 Jump Game (local best and global best method)