LeetCodeOJ.Jump Game II

试题请参见: https://leetcode.com/problems/jump-game-ii/

题目概述

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.)

解题思路

一看到最优解立即就想到了 BFS. 看完这题就联想到了 HDOJ 3278.

然而单纯的用 BFS 并不能 AC, 有一个丧心病狂的测试点[25000, 24999, 24998, ..., 1].

为了解决这个问题, 我们需要用一下贪心策略, 即我们每次走尽可能远的距离.

遇到的问题

TLE 了一次, 于是就引入了“贪心”策略.

源代码

#include 
#include 
#include 

class Solution {
public:
    int jump(std::vector<int>& nums) {
        // Initialize
        for ( int i = 0; i < nums.size(); ++ i ) {
            steps.push_back(0);
        }
        nodes.push(0);

        // BFS
        while ( !nodes.empty() ) {
            int currentNode = nodes.front();
            nodes.pop();

            int step = nums[currentNode];
            for ( int i = step; i >= 1; -- i ) {
                int nextNode = currentNode + i;

                if ( nextNode >= nums.size() ) {
                    continue;
                }
                if ( steps[nextNode] == 0 ) {
                    nodes.push(nextNode);
                }
                if ( steps[nextNode] == 0 ||
                     steps[nextNode] > steps[currentNode] + 1 ) {
                    steps[nextNode] = steps[currentNode] + 1;
                }

                // The last node is reached
                if ( nextNode == nums.size() - 1 ) {
                    return steps[nextNode];
                }
            }
        }
        return 0;
    }
private:
    std::vector<int> steps;
    std::queue<int> nodes;
};

int main() {
    std::vector<int> nums;

    Solution s;
    nums.push_back(2);
    nums.push_back(3);
    nums.push_back(1);
    nums.push_back(1);
    nums.push_back(4);

    std::cout << s.jump(nums) << std::endl;
    return 0;
}