題目如下:

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.)

分析如下：

看下面的圖來理解index的跳動情況

我的程式碼:

學習左耳朵耗子的程式碼。

// 10ms
class Solution {
public:
    int jump(int A[], int n) {
        if (n <= 1) return 0;
        int coverPro = 0;
        int steps = 0;
        for (int i = 0; i <= coverPro && coverPro < n;) {
            if (coverPro >= n-1) return steps;
            if (A[i] == 0) return -1;
            if (A[i] + i >= coverPro) {
                coverPro = A[i] + i;
                ++steps;
            }
            int maxDistance = 0;
            int nextStart = i + 1;
            for (int j = i + 1; j <= coverPro; ++j) {
                if(A[j] + j > maxDistance) {
                    maxDistance = A[j] + j;
                    nextStart = j;
                }
            }
            i = nextStart;
        }
        return steps;
    }
};
 

一個更加簡潔的答案，來自Kyle。

//10ms
class Solution {
public:
    int jump(int A[], int n) {
        if (n == 1) return 0;

        int step = 0;
        int index = 0;
        int distance = A[0];
        while (distance < n - 1) {
            ++step;
            int next_distance = distance;
            while (index < n && index <= distance) {
                next_distance = max(next_distance, index + A[index]);
                ++index;
            }
            if (next_distance <= distance) 
                return -1;
            distance = next_distance;
        }
        return step + 1;
    }
};