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.)
Note:
You can assume that you can always reach the last index.
Subscribe to see which companies asked this question
分析:
ret:目前为止的jump数
curRch:从A[0]进行ret次jump之后达到的最大范围
curMax:从0~i这i+1个A元素中能达到的最大范围
当curRch < i,说明ret次jump已经不足以覆盖当前第i个元素,因此需要增加一次jump,使之达到
记录的curMax。
ac代码:
class Solution {
public:
int jump(vector<int>& nums) {
int ret = 0;
int curMax = 0;
int curRch = 0;
int n=nums.size();
for(int i = 0; i < n; i ++)
{
if(curRch < i)
{
ret ++;
curRch = curMax;
}
curMax = max(curMax, nums[i]+i);
}
return ret;
}
};
本文介绍了一个算法问题,即如何在给定的非负整数数组中找到从第一个元素到达最后一个元素所需的最小跳跃次数。通过维护最大可达范围和当前可达范围等变量来优化跳跃路径。
2199

被折叠的 条评论
为什么被折叠?



