Given an array A
of integers, a ramp is a tuple (i, j)
for which i < j
and A[i] <= A[j]
. The width of such a ramp is j - i
.
Find the maximum width of a ramp in A
. If one doesn‘t exist, return 0.
Example 1:
Input: [6,0,8,2,1,5] Output: 4 Explanation: The maximum width ramp is achieved at (i, j) = (1, 5): A[1] = 0 and A[5] = 5.
Example 2:
Input: [9,8,1,0,1,9,4,0,4,1] Output: 7 Explanation: The maximum width ramp is achieved at (i, j) = (2, 9): A[2] = 1 and A[9] = 1.
Note:
2 <= A.length <= 50000
0 <= A[i] <= 50000
题意在解释里说的很清楚了,就看怎么想。
首先保证后面的数字比前面的是大于等于关系,这个排序可以搞定,然后到这个数为止,前面数字的位置里最小的数是谁,减去就好了
struct P{ int num; int pos; }H[50002]; bool cmp(P a,P b){ if(a.num == b.num){ return a.pos<b.pos; }else{ return a.num<b.num; } } class Solution { public: int maxWidthRamp(vector<int>& A) { int len = A.size(); for(int i=0;i<len;i++){ H[i].num = A[i]; H[i].pos = i; } sort(H,H+len,cmp); int Min = H[0].pos; int sum = -1; int cum = 0; for(int i=1;i<len;i++){ //cout<<H[i].pos<<" "<<Min<<endl; if(Min > H[i].pos){ Min = H[i].pos; }else{ cum = H[i].pos - Min; sum = max(sum,cum); } } if(sum == -1){ sum = 0; } return sum; } };
原文地址:https://www.cnblogs.com/yinghualuowu/p/10166554.html
时间: 2024-10-01 09:30:35