@@ -83,32 +83,245 @@ tags:
8383
8484<!-- solution:start -->
8585
86- ### 方法一
86+ ### 方法一:分类讨论
87+ 88+ 行和列都是回文的,那么对于任意 $i \in [ 0, m / 2)$ 和 $j \in [ 0, n / 2),ドル都需要满足 $\text{grid}[ i] [ j ] = \text{grid}[ m - i - 1] [ j ] = \text{grid}[ i] [ n - j - 1 ] = \text{grid}[ m - i - 1] [ n - j - 1 ] ,ドル要么都变成 0ドル,ドル要么都变成 1ドル,ドル变成 0ドル$ 的次数为 $c_0 = \text{grid}[ i] [ j ] + \text{grid}[ m - i - 1] [ j ] + \text{grid}[ i] [ n - j - 1 ] + \text{grid}[ m - i - 1] [ n - j - 1 ] ,ドル变成 1ドル$ 的次数为 $c_1 = 4 - c_0,ドル取两者的较小值,累加到答案中。
89+ 90+ 接下来,我们再讨论 $m$ 和 $n$ 的奇偶性:
91+ 92+ - 如果 $m$ 和 $n$ 都是偶数,那么直接返回答案;
93+ - 如果 $m$ 和 $n$ 都是奇数,那么最中间的格子只能是 0ドル,ドル因为题目要求 1ドル$ 的数目可以被 4ドル$ 整除;
94+ - 如果 $m$ 是奇数,而 $n$ 是偶数,那么我们需要考虑最中间的一行;
95+ - 如果 $m$ 是偶数,而 $n$ 是奇数,那么我们需要考虑最中间的一列。
96+ 97+ 对于后两种情况,我们需要统计最中间的一行或一列中对应位置不相同的格子对数 $\text{diff},ドル以及对应位置相同且为 1ドル$ 的格子个数 $\text{cnt1},ドル然后再分情况讨论:
98+ 99+ - 如果 $\text{cnt1} \bmod 4 = 0,ドル那么我们只需要将 $\text{diff}$ 个格子变成 0ドル$ 即可,操作次数为 $\text{diff}$;
100+ - 否则,说明 $\text{cnt1} = 2,ドル此时如果 $\text{diff} \lt 0,ドル我们可以将其中一个格子变成 1ドル,ドル使得 $\text{cnt1} = 4,ドル那么剩下的 $\text{diff} - 1$ 个格子变成 0ドル$ 即可,操作次数一共为 $\text{diff}$。
101+ - 否则,如果 $\text{diff} = 0,ドル我们就把 $\text{2}$ 个格子变成 0ドル,ドル使得 $\text{cnt1} \bmod 4 = 0,ドル操作次数为 $\text{2}$。
102+ 103+ 我们将操作次数累加到答案中,最后返回答案即可。
104+ 105+ 时间复杂度 $O(m \times n),ドル其中 $m$ 和 $n$ 分别是矩阵的行数和列数。空间复杂度 $O(1)$。
87106
88107<!-- tabs:start -->
89108
90109#### Python3
91110
92111``` python
93- 112+ class Solution :
113+ def minFlips (self grid : List[List[int ]]) -> int :
114+ m, n = len (grid), len (grid[0 ])
115+ ans = 0
116+ for i in range (m // 2 ):
117+ for j in range (n // 2 ):
118+ x, y = m - i - 1 , n - j - 1
119+ cnt1 = grid[i][j] + grid[x][j] + grid[i][y] + grid[x][y]
120+ ans += min (cnt1, 4 - cnt1)
121+ if m % 2 and n % 2 :
122+ ans += grid[m // 2 ][n // 2 ]
123+ diff = cnt1 = 0
124+ if m % 2 :
125+ for j in range (n // 2 ):
126+ if grid[m // 2 ][j] == grid[m // 2 ][n - j - 1 ]:
127+ cnt1 += grid[m // 2 ][j] * 2
128+ else :
129+ diff += 1
130+ if n % 2 :
131+ for i in range (m // 2 ):
132+ if grid[i][n // 2 ] == grid[m - i - 1 ][n // 2 ]:
133+ cnt1 += grid[i][n // 2 ] * 2
134+ else :
135+ diff += 1
136+ ans += diff if cnt1 % 4 == 0 or diff else 2
137+ return ans
94138```
95139
96140#### Java
97141
98142``` java
99- 143+ class Solution {
144+ public int minFlips (int [][] grid ) {
145+ int m = grid. length, n = grid[0 ]. length;
146+ int ans = 0 ;
147+ for (int i = 0 ; i < m / 2 ; ++ i) {
148+ for (int j = 0 ; j < n / 2 ; ++ j) {
149+ int x = m - i - 1 , y = n - j - 1 ;
150+ int cnt1 = grid[i][j] + grid[x][j] + grid[i][y] + grid[x][y];
151+ ans += Math . min(cnt1, 4 - cnt1);
152+ }
153+ }
154+ if (m % 2 == 1 && n % 2 == 1 ) {
155+ ans += grid[m / 2 ][n / 2 ];
156+ }
157+ 158+ int diff = 0 , cnt1 = 0 ;
159+ if (m % 2 == 1 ) {
160+ for (int j = 0 ; j < n / 2 ; ++ j) {
161+ if (grid[m / 2 ][j] == grid[m / 2 ][n - j - 1 ]) {
162+ cnt1 += grid[m / 2 ][j] * 2 ;
163+ } else {
164+ diff += 1 ;
165+ }
166+ }
167+ }
168+ if (n % 2 == 1 ) {
169+ for (int i = 0 ; i < m / 2 ; ++ i) {
170+ if (grid[i][n / 2 ] == grid[m - i - 1 ][n / 2 ]) {
171+ cnt1 += grid[i][n / 2 ] * 2 ;
172+ } else {
173+ diff += 1 ;
174+ }
175+ }
176+ }
177+ ans += cnt1 % 4 == 0 || diff > 0 ? diff : 2 ;
178+ return ans;
179+ }
180+ }
100181```
101182
102183#### C++
103184
104185``` cpp
105- 186+ class Solution {
187+ public:
188+ int minFlips(vector<vector<int >>& grid) {
189+ int m = grid.size(), n = grid[ 0] .size();
190+ int ans = 0;
191+ for (int i = 0; i < m / 2; ++i) {
192+ for (int j = 0; j < n / 2; ++j) {
193+ int x = m - i - 1, y = n - j - 1;
194+ int cnt1 = grid[ i] [ j ] + grid[ x] [ j ] + grid[ i] [ y ] + grid[ x] [ y ] ;
195+ ans += min(cnt1, 4 - cnt1);
196+ }
197+ }
198+ if (m % 2 == 1 && n % 2 == 1) {
199+ ans += grid[ m / 2] [ n / 2 ] ;
200+ }
201+ 202+ int diff = 0, cnt1 = 0;
203+ if (m % 2 == 1) {
204+ for (int j = 0; j < n / 2; ++j) {
205+ if (grid[m / 2][j] == grid[m / 2][n - j - 1]) {
206+ cnt1 += grid[m / 2][j] * 2;
207+ } else {
208+ diff += 1;
209+ }
210+ }
211+ }
212+ if (n % 2 == 1 ) {
213+ for (int i = 0; i < m / 2; ++i) {
214+ if (grid[i][n / 2] == grid[m - i - 1][n / 2]) {
215+ cnt1 += grid[i][n / 2] * 2;
216+ } else {
217+ diff += 1;
218+ }
219+ }
220+ }
221+ ans += cnt1 % 4 == 0 || diff > 0 ? diff : 2;
222+ return ans;
223+ }
224+ };
106225```
107226
108227#### Go
109228
110229``` go
230+ func minFlips (grid [][]int ) int {
231+ m , n := len (grid), len (grid[0 ])
232+ ans := 0
233+ 234+ for i := 0 ; i < m/2 ; i++ {
235+ for j := 0 ; j < n/2 ; j++ {
236+ x , y := m-i-1 , n-j-1
237+ cnt1 := grid[i][j] + grid[x][j] + grid[i][y] + grid[x][y]
238+ ans += min (cnt1, 4 -cnt1)
239+ }
240+ }
241+ 242+ if m%2 == 1 && n%2 == 1 {
243+ ans += grid[m/2 ][n/2 ]
244+ }
245+ 246+ diff , cnt1 := 0 , 0
247+ 248+ if m%2 == 1 {
249+ for j := 0 ; j < n/2 ; j++ {
250+ if grid[m/2 ][j] == grid[m/2 ][n-j-1 ] {
251+ cnt1 += grid[m/2 ][j] * 2
252+ } else {
253+ diff += 1
254+ }
255+ }
256+ }
257+ 258+ if n%2 == 1 {
259+ for i := 0 ; i < m/2 ; i++ {
260+ if grid[i][n/2 ] == grid[m-i-1 ][n/2 ] {
261+ cnt1 += grid[i][n/2 ] * 2
262+ } else {
263+ diff += 1
264+ }
265+ }
266+ }
267+ 268+ if cnt1%4 == 0 || diff > 0 {
269+ ans += diff
270+ } else {
271+ ans += 2
272+ }
273+ 274+ return ans
275+ }
276+ ```
111277
278+ #### TypeScript
279+ 280+ ``` ts
281+ function minFlips(grid : number [][]): number {
282+ const = grid .length ;
283+ const = grid [0 ].length ;
284+ let ans = 0 ;
285+ 286+ for (let i = 0 ; i < Math .floor (m / 2 ); i ++ ) {
287+ for (let j = 0 ; j < Math .floor (n / 2 ); j ++ ) {
288+ const = m - i - 1 ;
289+ const = n - j - 1 ;
290+ const = grid [i ][j ] + grid [x ][j ] + grid [i ][y ] + grid [x ][y ];
291+ ans += Math .min (cnt1 , 4 - cnt1 );
292+ }
293+ }
294+ 295+ if (m % 2 === 1 && n % 2 === 1 ) {
296+ ans += grid [Math .floor (m / 2 )][Math .floor (n / 2 )];
297+ }
298+ 299+ let diff = 0 ,
300+ cnt1 = 0 ;
301+ 302+ if (m % 2 === 1 ) {
303+ for (let j = 0 ; j < Math .floor (n / 2 ); j ++ ) {
304+ if (grid [Math .floor (m / 2 )][j ] === grid [Math .floor (m / 2 )][n - j - 1 ]) {
305+ cnt1 += grid [Math .floor (m / 2 )][j ] * 2 ;
306+ } else {
307+ diff += 1 ;
308+ }
309+ }
310+ }
311+ 312+ if (n % 2 === 1 ) {
313+ for (let i = 0 ; i < Math .floor (m / 2 ); i ++ ) {
314+ if (grid [i ][Math .floor (n / 2 )] === grid [m - i - 1 ][Math .floor (n / 2 )]) {
315+ cnt1 += grid [i ][Math .floor (n / 2 )] * 2 ;
316+ } else {
317+ diff += 1 ;
318+ }
319+ }
320+ }
321+ 322+ ans += cnt1 % 4 === 0 || diff > 0 ? diff : 2 ;
323+ return ans ;
324+ }
112325```
113326
114327<!-- tabs:end -->
0 commit comments