Leetcode 1020. Partition Array Into Three Parts With Equal Sum
题目：
Given an array A of integers, return true if and only if we can partition the array into three non-empty parts with equal sums.
Formally, we can partition the array if we can find indexes i+1 < j with (A[0] + A[1] + … + A[i] == A[i+1] + A[i+2] + … + A[j-1] == A[j] + A[j-1] + … + A[A.length - 1])
Example 1:
Input: [0,2,1,-6,6,-7,9,1,2,0,1]
Output: true
Explanation: 0 + 2 + 1 = -6 + 6 - 7 + 9 + 1 = 2 + 0 + 1
Example 2:
Input: [0,2,1,-6,6,7,9,-1,2,0,1]
Output: false
Example 3:
Input: [3,3,6,5,-2,2,5,1,-9,4]
Output: true
Explanation: 3 + 3 = 6 = 5 - 2 + 2 + 5 + 1 - 9 + 4
解析：其实就是前缀和的应用。找到l，r即可。
代码：
class Solution
{
public:
bool canThreePartsEqualSum(vector<int>& A)
{
int sum = 0;
int size = A.size();
vector<int> dp(size,0);
for(int i = 0; i < size; i++)
{
if(i==0) dp[0] = A[i];
else dp[i] = dp[i-1] + A[i];
sum += A[i];
}
if(sum%3!=0) return false;
int l=-1,r=-1;
for(int i = 0; i<size;i++)
{
if(dp[i] == sum/3)
{
l = i;
break;
}
}
if(l==-1) return false;
for(int i = l+1;i<size;i++)
{
if(dp[i]-dp[l] == sum/3)
{
r = i;
break;
}
}
if(r==-1) return false;
if(dp[size-1]-dp[r]==sum/3) return true;
else return false;
}
};
Leetcode 1022. Smallest Integer Divisible by K
Given a positive integer K, you need find the smallest positive integer N such that N is divisible by K, and N only contains the digit 1.
Return the length of N. If there is no such N, return -1.
Example 1:
Input: 1
Output: 1
Explanation: The smallest answer is N = 1, which has length 1.
Example 2:
Input: 2
Output: -1
Explanation: There is no such positive integer N divisible by 2.
Example 3:
Input: 3
Output: 3
Explanation: The smallest answer is N = 111, which has length 3.
Note:
1 <= K <= 10^5
题意：是否存在只由1构成的数，整数K。
解析：用模拟除法的方法去不停的mod，并且假装后面有很多1。
代码：
class Solution
{
public:
int smallestRepunitDivByK(int K)
{
int cnt = 0;
int value = 0;
int i;
for(i = 0; i < 1e6; i++)
{
int value = (value*10+1)%K;
++cnt;
if(value==0) break;
}
if(i==1e6) return -1;
return cnt;
}
};
Leetcode 1021. Best Sightseeing Pair
Given an array A of positive integers, A[i] represents the value of the i-th sightseeing spot, and two sightseeing spots i and j have distance j - i between them.
The score of a pair (i < j) of sightseeing spots is (A[i] + A[j] + i - j) : the sum of the values of the sightseeing spots, minus the distance between them.
Return the maximum score of a pair of sightseeing spots.
Example 1:
Input: [8,1,5,2,6]
Output: 11
Explanation: i = 0, j = 2, A[i] + A[j] + i - j = 8 + 5 + 0 - 2 = 11
Note:
2 <= A.length <= 50000
1 <= A[i] <= 1000
题意：就是给很多个景点，然后两个景点之间有个衡量标准，求一个最大值。
解析：我是左扫一次，比如到第i点，我得到的是从[0,i)的一个左闭右开的区间的最大值，而从右往前扫，我维护的是一个[i，size-1] 左右皆闭合的区间。这样最后扫一次，就可以得到最大值了。
代码：
class Solution
{
public:
int maxScoreSightseeingPair(vector<int>& A)
{
int size = A.size();
int ans = -1;
vector<int> l_max(size,0);
vector<int> r_max(size,0);
l_max[1] = max(l_max[0],A[0]+0);
for(int i = 2;i<size;i++)
l_max[i] = max(l_max[i-1],A[i-1]+i-1);
r_max[size-1] = A[size-1]-(size-1);
r_max[size-2] = max(r_max[size-1],A[size-2]-(size-2));
for(int i = size-3;i>=1;i--)
r_max[i] = max(r_max[i+1],A[i]-i);
for(int i = 0;i<size;i++)
{
//cout << l_max[i] << " " << r_max[i] <<endl;
ans = max(l_max[i]+r_max[i],ans);
}
return ans;
}
};
Leetcode 1023. Binary String With Substrings Representing 1 To N
Given a binary string S (a string consisting only of ‘0’ and '1’s) and a positive integer N, return true if and only if for every integer X from 1 to N, the binary representation of X is a substring of S.
Example 1:
Input: S = “0110”, N = 3
Output: true
Example 2:
Input: S = “0110”, N = 4
Output: false
Note:
1 <= S.length <= 1000
1 <= N <= 10^9
题意：就是给一个母串，给一个数N，看是否从1-N的所有的数的二进制表示都是母串的子串。
这个题没想到是直接暴力。。。
class Solution
{
public:
bool queryString(string S, int N)
{
for(int i = 1;i<=N;i++)
{
string str = "";
int x = i;
while(x)
{
str += x%2+'0';
x /= 2;
}
reverse(str.begin(), str.end());
if(S.find(str)==string::npos) return false;
}
return true;
}
};