LEETCODE: Jump Game II

原创于 2014-12-18 17:25:34 发布 · 382 阅读

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#LEETCODE #算法 #贪心算法

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本文介绍了一个使用贪心算法解决的问题——如何在给定数组中从第一个元素跳到最后一个元素所需的最少跳跃次数。通过实例演示了算法的具体实现过程。

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.)

贪心算法：

class Solution {
public:
    int jump(int A[], int n) {
        if(n <= 1) return 0;
        int max = A[0];
        int step = 1;
        int pos = 0;
        while(max < n - 1) {
            int lastmax = max;
            for(int ii = pos + 1; ii < lastmax + 1; ii ++) {
                if(max < A[ii] + ii) {
                    max = A[ii] + ii;
                }
            }
            step ++;
        }
        return step;
    }
};