方法一：枚举
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| class Solution {
public int countSymmetricIntegers(int low, int high) {
int res = 0;
for (int i = low; i <= Math.min(high, 99); ++i) {
int l = i % 10, r = i / 10;
if (l == r)
++res;
}
if (high > 1000) {
for (int i = Math.max(low, 1000); i <= Math.min(high, 10000); ++i) {
int pre = 0, suff = 0;
int l = i;
pre += (l % 10);
l /= 10;
pre += (l % 10);
int r = i / 100;
suff += (r % 10);
r /= 10;
suff += (r % 100);
if (pre == suff)
++res;
}
}
return res;
}
}
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方法一：贪心
从后往前找，找到子序列25， 50， 75， 00，统计最少需要删除的个数；如果没找到，那么如果字符串中没有0，那么全部删除；如果有0，则删除n-1个字符
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| class Solution {
public int minimumOperations(String s) {
int n = s.length();
int cnt0 = 0;
for (int i = 0; i < n; ++i) {
if (s.charAt(i) == '0')
cnt0++;
}
int res = 101;
int flag = 0;
int cnt = 0;
for (int i = n - 1; i >= 0; --i) {
char ch = s.charAt(i);
if (flag == 0) {
if (ch == '0')
flag = 1;
else
++cnt;
} else if (flag == 1) {
if (ch == '0' || ch == '5')
flag = 2;
else
++cnt;
} else
break;
}
if (flag == 2)
res = Math.min(res, cnt);
flag = 0;
cnt = 0;
for (int i = n - 1; i >= 0; --i) {
char ch = s.charAt(i);
if (flag == 0) {
if (ch == '5')
flag = 1;
else
++cnt;
} else if (flag == 1) {
if (ch == '2' || ch == '7')
flag = 2;
else
++cnt;
} else
break;
}
if (flag == 2)
res = Math.min(res, cnt);
if (res == 101) {
res = cnt0 == 0 ? n : n - 1;
}
return res;
}
}
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方法一：前缀和 + HashMap + 公式转换
类似两数之和
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| public long countInterestingSubarrays(List<Integer> nums, int modulo, int k) {
Map<Integer, Integer> map = new HashMap<>();
map.put(0, 1);
long res = 0;
int s = 0;
for (int x : nums) {
if (x % modulo == k)
s = (s + 1) % modulo;
res += map.getOrDefault((s - k + modulo) % modulo, 0);
map.put(s, map.getOrDefault(s, 0) + 1);
}
return res;
}
|
