lc3047
遍历所有矩形对，计算每对矩形的交集区域，取++交集的宽高min++作为可能的正方形边长，记录最大边长并返回其平方值。
class Solution {
public:
long long largestSquareArea(vector<vector<int>>& bottomLeft, vector<vector<int>>& topRight) {
int max_side = 0;
for (int i = 0; i < bottomLeft.size(); i++) {
auto& b1 = bottomLeft[i];
auto& t1 = topRight[i];
for (int j = 0; j < i; j++) {
auto& b2 = bottomLeft[j];
auto& t2 = topRight[j];
int width = min(t1[0], t2[0]) - max(b1[0], b2[0]);
int height = min(t1[1], t2[1]) - max(b1[1], b2[1]);
++int side = min(width, height);
max_side = max(max_side, side);++
}
}
return 1LL * max_side * max_side;
}
};
单调性cut
class Solution {
public:
long long largestSquareArea(vector<vector<int>>& bottomLeft, vector<vector<int>>& topRight) {
int max_side = 0;
for (int i = 0; i < bottomLeft.size(); i++) {
auto& b1 = bottomLeft[i];
auto& t1 = topRight[i];
++if (t1[0] - b1[0] <= max_side || t1[1] - b1[1] <= max_side)
continue;++ // 最优性剪枝：max_side 不可能变大
for (int j = 0; j < i; j++) {
auto& b2 = bottomLeft[j];
auto& t2 = topRight[j];
int width = min(t1[0], t2[0]) - max(b1[0], b2[0]); // 右上横坐标 - 左下横坐标
int height = min(t1[1], t2[1]) - max(b1[1], b2[1]); // 右上纵坐标 - 左下纵坐标
int side = min(width, height);
max_side = max(max_side, side);
}
}
return 1LL * max_side * max_side;
}
};
lc3796
正反两次扫描
class Solution {
public:
int findMaxVal(int n, vector<vector<int>>& restrictions, vector<int>& diff) {
vector<int> max_val(n, INT_MAX);
for (auto& r : restrictions)
max_val[r[0]] = r[1];
vector<int> a(n);
for (int i = 0; i < n - 1; i++) {
a[i + 1] = min(a[i] + diff[i], max_val[i + 1]);
}
for (int i = n - 2; i > 0; i--) {
a[i] = min(a[i], a[i + 1] + diff[i]);
}
return ranges::max(a);
}
};
lc3795
hash+滑窗
class Solution {
public:
int minLength(vector<int>& nums, int k) {
unordered_map<int, int> cnt;
int sum = 0;
int left = 0;
int ans = INT_MAX;
for (int i = 0; i < nums.size(); i++) {
// 1. 入
int x = nums[i];
cnt[x]++;
if (cnt[x] == 1)
sum += x;
while (sum >= k) {
// 2. 更新答案
ans = min(ans, i - left + 1);
// 3. 出
int out = nums[left];
cnt[out]--;
if (cnt[out] == 0)
sum -= out;
left++;
}
}
return ans == INT_MAX ? -1 : ans;
}
};