面试笔试杂项积累-leetcode 131-135

131.131-Palindrome Partitioning-Difficulty: Medium

Given a string s, partition s such that every substring of the partition is a palindrome.

Return all possible palindrome partitioning of
s.

For example, given s =

"aab"

,

Return

[
["aa","b"],
["a","a","b"]
]

思路

给予一个string，找到所有回文的部分

动态规划

public class Solution {
public IList<IList<string>> Partition(string s) {
IList<IList<string>> res =new List<IList<string>>();
if (s=="") {
res.Add(new List<string>());
return res;
}
for (int i = 0; i < s.Length; i++) {
if (isPalindrome(s, i + 1)) {
foreach (List<string> list in Partition(s.Substring(i+1))) {
list.Insert(0, s.Substring(0, i + 1));
res.Add(list);
}
}
}
return res;
}
public bool isPalindrome(string s, int n) {
for (int i = 0; i < n / 2; i++) {
if (s[i] != s[n - i - 1])
return false;
}
return true;
}

}

132.132-Palindrome Partitioning II-Difficulty: Hard

Given a string s, partition s such that every substring of the partition is a palindrome.

Return the minimum cuts needed for a palindrome partitioning of
s.

For example, given s =

"aab"

,

Return

1

since the palindrome partitioning

["aa","b"]

could be produced using 1 cut.

思路

找到最少分割的回文，返回分割次数

动态规划

This can be solved by two points:

cut[i]

is the minimum of

cut[j - 1] + 1 (j <= i)

, if

[j, i]

is palindrome.
If

[j, i]

is palindrome,

[j + 1, i - 1]

is palindrome, and

c[j] == c[i]

.
The 2nd point reminds us of using dp (caching).

a   b   a   |   c  c
j  i
j-1  |  [j, i] is palindrome
cut(j-1) +  1

参考：
https://leetcode.com/discuss/76411/easiest-java-dp-solution-97-36%25
public class Solution {
public int MinCut(string s) {
char[] c = s.ToCharArray();
int n = c.Length;
int[] cut = new int
;
bool[,] pal = new bool[n,n];

for(int i = 0; i < n; i++) {
int min = i;
for(int j = 0; j <= i; j++) {
if(c[j] == c[i] && (j + 1 > i - 1 || pal[j + 1,i - 1])) {
pal[j,i] = true;
min = j == 0 ? 0 : Math.Min(min, cut[j - 1] + 1);
}
}
cut[i] = min;
}
return cut[n - 1];
}
}

134.134-Gas Station-Difficulty: Medium

There are N gas stations along a circular route, where the amount of gas at station
i is

gas[i]

.

You have a car with an unlimited gas tank and it costs

cost[i]

of gas to travel from station i to its next station (i+1). You begin the journey with an empty tank at one of the gas stations.

Return the starting gas station's index if you can travel around the circuit once, otherwise return -1.

Note:

The solution is guaranteed to be unique.

思路

给与一个数组gas[]代表某个位置(i)的瓦斯，一个数组cost[]代表从当前位置(i)到下一个位置(i+1)需要消耗多少瓦斯，找到一条能从开始走到结尾的一条路线，返回起始位置

从每个可行起始位置开始，先要判断哪个位置可行，判断方法是gas[i]>=cost[i]这个是必须的，得有至少相等的gas到下一个位置。然后再遍历如果cost大于已有gas则丢弃此起始节点从下一个再开

public class Solution {
public int CanCompleteCircuit(int[] gas, int[] cost) {
int start = 0;
int have_gas = 0;
int j = gas.Length - 1;
for (int i = 0; i < gas.Length; i++)
{
have_gas = 0;
j = gas.Length - 1;
if (gas[i] >= cost[i])
{
start = i;
have_gas = 0;
while (j > -1)
{
have_gas += gas[start] - cost[start];
if (have_gas < 0)
break;
++start;
if (j == i)
start = 0;
--j;
}
if (j == -1)
return i;
}
}
return -1;
}
}

135.135-Candy-Difficulty: Hard

There are N children standing in a line. Each child is assigned a rating value.

You are giving candies to these children subjected to the following requirements:

Each child must have at least one candy.
Children with a higher rating get more candies than their neighbors.
What is the minimum candies you must give?

思路

分糖，

每个孩子至少要有一个糖，高优先级的孩子的糖要比左右邻居的要多

从左到右分一遍，从右到左分一遍，就是要照顾到这个左右啊。。。

动态规划

public class Solution {
public int Candy(int[] ratings) {
int n = ratings.Length;
int[] candy = new int
;
for (int i = 0; i < n; i++)
candy[i] = 1;
for (int i = 1; i < n; ++i)
{
candy[i] = ratings[i] <= ratings[i - 1] ? 1 : candy[i - 1] + 1;
}
for (int i = n - 2; i > -1; --i)
{
candy[i] = ratings[i] <= ratings[i + 1] ? candy[i] : Math.Max(candy[i], candy[i + 1] + 1);
}
for (int i = 1; i < n; i++)
candy[0] += candy[i];
return candy[0];
}
}