"Greedy algorithm, dynamic planning" Jump Game II

Title: Leetcode

Jump Game II 

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.

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 and then steps to the last 3 index.)

Analysis:

Should consider unable to reach the end of the situation, some online answers do not consider this.

Idea 1: Greedy algorithm, time complexity O (n), Space complexity O (1)

int jump (int a[], int n) {int ret = 0;int Curmax = 0;int Currch = 0;for (int i = 0; i < n; i++) {if (Currch < i) {ret+ +;currch = curmax;if (Currch < i) return-1;//return-1 means unable to reach the end point}curmax = Max (Curmax, a[i] + i);} return ret;}

"Greedy algorithm, dynamic planning" Jump Game II