[leetcode 45] Jump Game II

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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跃到最后一个元素所需的最少跳跃次数。通过定义活动范围并不断更新最远可达位置来实现算法。

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.)

思路：贪心。 [left, right]为在当前位置的活动范围，找到在这个范围内能跳到最远的位置。

class Solution {
public:
    int jump(int A[], int n) {
        if (n < 2) {
            return 0;
        }
        int step = 0;
        int left = 0;
        int right = 0;
        while (left <= right) {
            step++;
            int old_right = right;
            for (int i = left; i <= old_right; i++) {
                int new_right = i + A[i];
                if (new_right >= n - 1) {
                    return step;
                }
                if (new_right > right) {
                    right = new_right;
                }
            }
            left = old_right + 1;
        }
        return 0;
    }
};