LeetCode——jump-game-ii

Question

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.)

Solution

1. 贪心算法。

2. 跳动范围：[start, end]，在这个跳动范围内，能跳最远记作far。一旦当前点到达end，那么会突发下一次跳跃(也就是在end或者end之前就必须跳了)，当然是跳到最远的地方去，所以把跳动范围的end设置为far即可。

3. 如果到达终点就结束

Code

class Solution {
public:
    int jump(int A[], int n) {
        int res = 0;
        if (n <= 1)
            return 0;
        int start = 0, end = 0, far = 0;
        for (int i = 0; i < n; i++) {
            far = max(far, A[i] + i);
            // end，或者end之前肯定会触发跳动的
            if (i == end) {
				res++;
                end = far;
                // 达到终点就退出
                if (end >= n - 1)
                    break;
            }
        }
        return res;
    }
};