力扣算法题—045跳跃游戏二

#include "000库函数.h"


//考虑当前最远能到什么地方，例如2, 3, 1, 1, 4，
//首先只考虑a[0] = 2，即最远可以到a[2]，然后从1到2中找下一个可到的最远点，
//即a[1]可以到达a[4]，此时找到结果，步数记录为2。若接着考虑，
//下一次应该从3 - 4里面找一个最远即a[4]可达a[8]（4 + 4）, 
//再下一次从5 - 8中找最远
//20ms
class Solution {
public:
    int jump(vector<int>& nums) {
        if (nums.size() < 2)return 0;
        int min = 0;
        int a = 0, b = 0 + nums[0];//第一步和下一步的坐标
        while (a < nums.size() - 1) {
            min++;    
            int s = MAX(nums, a + 1, b);    //在a,b中找到最远距离
            a = b;
            b = s;
        }
        return min;
    }
    int MAX(vector<int>& nums, int s, int t) {
        int max = s + nums[s];
        for (; s < nums.size() && s <= t; ++s) {
            if (max < s + nums[s])
                max = s + nums[s];
        }
        return max;
    }
};

//博客解法  20ms
//解法的思想一样，找到跳的最远的那个
class Solution {
public:
    int jump(vector<int>& nums) {
        int res = 0, n = nums.size(), i = 0, cur = 0;
        while (cur < n - 1) {
            ++res;
            int pre = cur;
            for (; i <= pre; ++i) {
                cur = max(cur, i + nums[i]);
            }
            if (pre == cur) return -1; // May not need this
        }
        return res;
    }
};


class Solution {
public:
    int jump(vector<int>& nums) {
        int res = 0, n = nums.size(), last = 0, cur = 0;
        for (int i = 0; i < n - 1; ++i) {
            cur = max(cur, i + nums[i]);
            if (i == last) {
                last = cur;
                ++res;
                if (cur >= n - 1) break;
            }
        }
        return res;
    }
};
void T045() {
    Solution s;
    vector<int>v;
    v = { 5,5,3,2,1,0,2,3,3,10,0,0 };
    cout << s.jump(v) << endl << "**************" << endl;
    v = { 2,0,2,4,6,0,0,3};
    cout << s.jump(v) << endl << "**************" << endl;
    v = { 2,3,0,1,4 };
    cout << s.jump(v) << endl << "**************" << endl;
    v = { 2,3,1,1,4 };
    cout << s.jump(v) << endl << "**************" << endl;
}