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徐长贺/Leetcode

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_672.java 2.53 KB
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stevesun 提交于 8年前 . make checkstyle happy
package com.fishercoder.solutions;
import com.fishercoder.common.utils.CommonUtils;
/**
* 672. Bulb Switcher II
* There is a room with n lights which are turned on initially and 4 buttons on the wall.
* After performing exactly m unknown operations towards buttons,
* you need to return how many different kinds of status of the n lights could be.
* Suppose n lights are labeled as number [1, 2, 3 ..., n], function of these 4 buttons are given below:
Flip all the lights.
Flip lights with even numbers.
Flip lights with odd numbers.
Flip lights with (3k + 1) numbers, k = 0, 1, 2, ...
Example 1:
Input: n = 1, m = 1.
Output: 2
Explanation: Status can be: [on], [off]
Example 2:
Input: n = 2, m = 1.
Output: 3
Explanation: Status can be: [on, off], [off, on], [off, off]
Example 3:
Input: n = 3, m = 1.
Output: 4
Explanation: Status can be: [off, on, off], [on, off, on], [off, off, off], [off, on, on].
Note: n and m both fit in range [0, 1000].
*/
public class _672 {
public static class Solution1 {
public int flipLights(int n, int m) {
if (m < 1) {
return 1;
}
int[][] dp = new int[m][n];
for (int i = 0; i < m; i++) {
for (int j = 0; j < n; j++) {
if (j == 0) {
dp[i][j] = 2;
} else if (i == 0 && j == 1) {
dp[i][j] = 3;
} else if (i == 0) {
dp[i][j] = 4;
} else if (i == 1 && j == 1) {
dp[i][j] = 4;
} else if (i == 1 && j > 1) {
dp[i][j] = 7;
} else if (j == 1) {
dp[i][j] = 4;
} else if (i == 1) {
dp[i][j] = 7;
} else {
dp[i][j] = 8;
}
}
}
CommonUtils.print2DIntArray(dp);
return dp[m - 1][n - 1];
}
}
public static class Solution2 {
public int flipLights(int n, int m) {
if (n == 1 && m > 0) {
return 2;
} else if (n == 2 && m == 1) {
return 3;
} else if ((n > 2 && m == 1) || (n == 2 && m > 1)) {
return 4;
} else if (n > 2 && m == 2) {
return 7;
} else if (n > 2 && m > 2) {
return 8;
} else {
return 1;
}
}
}
}
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