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导航网站的网站地图怎么做,广州市司法职业学校,泰安市卓创网络科技有限公司,网站建设要固定ip网络吗【题目描述】 在给定的一个nums数组中#xff0c;nums[i]表示从当前i位置最多可以向后跳跃nums[i]个位置。问跳跃到最后 数组最后一个元素的最少跳跃次数#xff1f;#xff1f;#xff1f; 【贪心】 首先#xff0c;需要注意的是#xff0c;题目的意思是最远跳跃nums[…【题目描述】在给定的一个nums数组中nums[i]表示从当前i位置最多可以向后跳跃nums[i]个位置。问跳跃到最后 数组最后一个元素的最少跳跃次数【贪心】首先需要注意的是题目的意思是最远跳跃nums[i]个位置而不是必须跳跃mun[i]个位置。以一个例子来看nums[2,3,1,1,4,2,6,7,1]。nums[0]2所以可以从0位置最多向后跳跃2个位置所以最远可以跳跃到1也可以选择跳跃到3。nums[2,3,1,1,4,2,6,7,1]。还是以这个数组为例。第一次的起跳位置是数组下标为0的位置。经过一次跳跃之后可以到达3或者1此时的位置是第二次起跳位置也就是说第二次起跳的位置是3或者1。同理假设3为第二次起跳的位置 那么可以到达【114】假设1为第二次起跳的位置可以到达【1】。两者取并集可以得到从第二次起跳位置起跳可以跳跃到的区间。后面也是同理配图从上图我们可以发现二次起跳的位置【31】和三次起跳的位置【114】这两个区间存在交集。而题目中求的是最少跳跃次数所以可以通过二次跳跃可以到达的位置不会选择跳跃三次。那么如何判断跳跃到数组的最后一个元素了对于每一个跳跃区间我们可以使用两个变量来维护这两个区间的开始位置和结束位置【leftright】比如 对于二次起跳的位置上图中蓝色部分的线对应的区间就是【12】。当right超出数组的长度就表示可以跳跃到数组的结尾位置了。注意这个【leftright】区间存储的是下标不是元素。维护【left,right】区间从上图可以发现第三次起跳的【leftright】其中left可以通过上一次起跳的【leftright】的right1得来。而right就需要遍历第二次起跳的【leftright】找出最远的距离。【代码】只需遍历数组一遍时间复杂度为O(N)代码语言javascriptAI代码解释class Solution { public: int jump(vectorint nums) { int nnums.size(); int maxpos0;//存储下一次起跳的最远位置也就是下一个right int ret0;//记录最终结果 int left0,right0; while(leftright) { if(maxposn-1)//已经可以跳跃到数组结尾 { return ret; } //遍历当前[left,right]更新下一次的[left,right] for(int ileft;iright;i) { maxposmax(maxpos,nums[i]i); } leftright1; rightmaxpos; ret;//跳跃次数 } return ret; } };2跳跃游戏I题目连接55. 跳跃游戏 - 力扣LeetCode本题与上体思路一模一样只是返回值修改以下即可。代码语言javascriptAI代码解释class Solution { public: bool canJump(vectorint nums) { //贪心 int nnums.size(); int left0,right0,ret0; int maxPos0;//记录下一个最远跳到的位置 while(leftright) { if(maxPosn-1) return true; //跟新maxPos for(int ileft;iright;i) maxPosmax(maxPos,nums[i]i); //更新区间 leftright1; rightmaxPos; ret; } return false; } };3加油站题目连接134. 加油站 - 力扣LeetCode【题目描述】题目简单的理解n个加油站每个加油站有gas[i]升汽油。刚开始汽车的油箱为空。从第i个加油站行驶到第i1个加油站需要消耗cost[i]升汽油。问是否存在一个加油站是该汽车可以环绕n个加油占一周回到起始加油站返回该加油站的下标。【暴力枚举算法】我们可以枚举每个加油站位置看看是否可以环绕一周回到起始位置。从第i个加油站到达第i1个加油站需要消耗cost[i]升汽油而第i个加油站可以获得gas[i]升汽油所以汽车要想从第i个加油站行驶到第i1个加油站必须保证gas[i]cos[i]。我们可以将gas[i]-cos[i]的值保存于一个diff数组中这个数组不需要定义出来只是为了好理解如果从第i个加油站开始走那么mixup满足gas[i]cos[i]也就是diff[i]0。以题目中给的示例为例当我们选中一个位置作为起点从当前位置再次枚举步数0~n-以为可能下标会越界所以需要在对数组长度取模。【暴力代码】会超时时间复杂度ON^2代码语言javascriptAI代码解释class Solution { public: int canCompleteCircuit(vectorint gas, vectorint cost) { int n gas.size(); //枚举起点 for(int i0;in;i) { int rest0;//剩余汽油 for(int step0;stepn;step)//枚举步数 { int index(stepi)%n;//求出走step后的下标 restgas[index]-cost[index]; if(rest0) break; } if(rest0) return i; } return -1; } };【贪心优化】代码语言javascriptAI代码解释class Solution { public: int canCompleteCircuit(vectorint gas, vectorint cost) { int n gas.size(); //枚举起点 for(int i0;in;i) { int rest0;//剩余汽油 int step0; for(;stepn;step)//枚举步数 { int index(stepi)%n;//求出走step后的下标 restgas[index]-cost[index]; if(rest0) break; } if(rest0) return i; istep;//循环会完成1工作 } return -1; } 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