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class Solution {
public:
int minFlips(vector<vector<int>>& grid) {
}
};
java 解法, 执行用时: 3 ms, 内存消耗: 93.7 MB, 提交时间: 2024-08-10 12:07:11
class Solution {
public int minFlips(int[][] a) {
int ans = 0;
int m = a.length;
int n = a[0].length;
for (int i = 0; i < m / 2; i++) {
for (int j = 0; j < n / 2; j++) {
int cnt1 = a[i][j] + a[i][n - 1 - j] + a[m - 1 - i][j] + a[m - 1 - i][n - 1 - j];
ans += Math.min(cnt1, 4 - cnt1); // 全为 1 或全为 0
}
}
if (m % 2 > 0 && n % 2 > 0) {
// 正中间的数必须是 0
ans += a[m / 2][n / 2];
}
int diff = 0, cnt1 = 0;
if (m % 2 > 0) {
// 统计正中间这一排
for (int j = 0; j < n / 2; j++) {
if (a[m / 2][j] != a[m / 2][n - 1 - j]) {
diff++;
} else {
cnt1 += a[m / 2][j] * 2;
}
}
}
if (n % 2 > 0) {
// 统计正中间这一列
for (int i = 0; i < m / 2; i++) {
if (a[i][n / 2] != a[m - 1 - i][n / 2]) {
diff++;
} else {
cnt1 += a[i][n / 2] * 2;
}
}
}
return ans + (diff > 0 ? diff : cnt1 % 4);
}
}
cpp 解法, 执行用时: 423 ms, 内存消耗: 189.2 MB, 提交时间: 2024-08-10 12:06:58
class Solution {
public:
int minFlips(vector<vector<int>>& a) {
int m = a.size(), n = a[0].size(), ans = 0;
for (int i = 0; i < m / 2; i++) {
for (int j = 0; j < n / 2; j++) {
int cnt1 = a[i][j] + a[i][n - 1 - j] + a[m - 1 - i][j] + a[m - 1 - i][n - 1 - j];
ans += min(cnt1, 4 - cnt1); // 全为 1 或全为 0
}
}
if (m % 2 && n % 2) {
// 正中间的数必须是 0
ans += a[m / 2][n / 2];
}
int diff = 0, cnt1 = 0;
if (m % 2) {
// 统计正中间这一排
for (int j = 0; j < n / 2; j++) {
if (a[m / 2][j] != a[m / 2][n - 1 - j]) {
diff++;
} else {
cnt1 += a[m / 2][j] * 2;
}
}
}
if (n % 2) {
// 统计正中间这一列
for (int i = 0; i < m / 2; i++) {
if (a[i][n / 2] != a[m - 1 - i][n / 2]) {
diff++;
} else {
cnt1 += a[i][n / 2] * 2;
}
}
}
return ans + (diff ? diff : cnt1 % 4);
}
};
python3 解法, 执行用时: 238 ms, 内存消耗: 56.1 MB, 提交时间: 2024-08-10 12:06:41
class Solution:
def minFlips(self, a: List[List[int]]) -> int:
ans = 0
m, n = len(a), len(a[0])
for i in range(m // 2):
row, row2 = a[i], a[-1 - i]
for j in range(n // 2):
cnt1 = row[j] + row[-1 - j] + row2[j] + row2[-1 - j]
ans += min(cnt1, 4 - cnt1) # 全为 1 或全为 0
if m % 2 and n % 2:
# 正中间的数必须是 0
ans += a[m // 2][n // 2]
diff = cnt1 = 0
if m % 2:
# 统计正中间这一排
row = a[m // 2]
for j in range(n // 2):
if row[j] != row[-1 - j]:
diff += 1
else:
cnt1 += row[j] * 2
if n % 2:
# 统计正中间这一列
for i in range(m // 2):
if a[i][n // 2] != a[- 1 - i][n // 2]:
diff += 1
else:
cnt1 += a[i][n // 2] * 2
return ans + (diff if diff else cnt1 % 4)
golang 解法, 执行用时: 290 ms, 内存消耗: 25.2 MB, 提交时间: 2024-08-10 12:06:26
func minFlips(a [][]int) (ans int) {
m, n := len(a), len(a[0])
for i, row := range a[:m/2] {
row2 := a[m-1-i]
for j, x := range row[:n/2] {
cnt1 := x + row[n-1-j] + row2[j] + row2[n-1-j]
ans += min(cnt1, 4-cnt1) // 全为 1 或全为 0
}
}
if m%2 > 0 && n%2 > 0 {
// 正中间的数必须是 0
ans += a[m/2][n/2]
}
diff, cnt1 := 0, 0
if m%2 > 0 {
// 统计正中间这一排
row := a[m/2]
for j, x := range row[:n/2] {
if x != row[n-1-j] {
diff++
} else {
cnt1 += x * 2
}
}
}
if n%2 > 0 {
// 统计正中间这一列
for i, row := range a[:m/2] {
if row[n/2] != a[m-1-i][n/2] {
diff++
} else {
cnt1 += row[n/2] * 2
}
}
}
if diff > 0 {
ans += diff
} else {
ans += cnt1 % 4
}
return
}
