[leetcode] Jump Game II

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本文介绍了一个算法问题——跳远游戏II，旨在找到从数组起始位置到达末尾所需的最少跳跃次数。通过分析每个元素作为最大跳跃长度，采用贪心策略选择下一步的最佳跳跃位置，最终实现用最少步骤完成目标。

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From : https://leetcode.com/problems/jump-game-ii/

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.)

class Solution {
public:
    int jump(vector<int>& nums) {
        int stp = 0, len=nums.size();
        if(len <= 1) return 0;
        for(int i=0; i<len;) {
            if(nums[i]) {
                ++stp;
            }
            int idx = i+1, far = 0, n=nums[i]+i;
            if(n+1 >= len) return stp;
            for(int j=idx; j<=n && j<len; ++j) {
                if(far < j+nums[j]) {
                    idx = j;
                    far = j+nums[j];
                }
            }
            i = idx;
        }
        return stp;
    }
};

public class Solution {
    public int jump(int[] nums) {
        int steps = 0, far = 0;
        for(int i=0, last=nums.length-1; i<last && nums[i]>0;) {
            far=i+nums[i]; //最远far
            if(far >= last) { // 已经能跳到了，那么再条一次
            	return steps+1;
            }
            for(int j=i,curLast=far; j<=curLast; ++j) {
                if(j+nums[j] > far) {
                    far = j+nums[j];
                    i = j;
                }
                if(far >= last) {
                    break;
                }
            }
            ++steps;//这一条
        }
        return steps;
    }
}