leetcode-474. 一和零

原始思路

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class Solution {
public int findMaxForm(String[] strs, int m, int n) {
int n1 = strs.length;
int[][][] dp = new int[n1+1][m+1][n+1];
for (int i = 1; i <= strs.length; i++) {
int[] arr = getZeorAndOne(strs[i-1]);
int zero = arr[0];
int one = arr[1];
for (int j = 0; j <= m; j++) {
for (int k = 0; k <= n; k++) {
dp[i][j][k] = dp[i-1][j][k];
if (zero <= j && one <= k){
dp[i][j][k] = Math.max(dp[i][j][k], dp[i-1][j - zero][k - one] + 1);
}
}
}
}
return dp[n1][m][n];
}

public int[] getZeorAndOne(String str){
int[] arr = new int[2];
int i = 0;
int j = 0;
for (int k = 0; k < str.length(); k++) {
if (str.charAt(k) == '0'){
i++;
}else {
j++;
}
}
arr[0] = i;
arr[1] = j;
return arr;
}
}

状态压缩

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class Solution {
public int findMaxForm(String[] strs, int m, int n) {
int n1 = strs.length;
int[][] dp = new int[m+1][n+1];
for (int i = 1; i <= strs.length; i++) {
int[] arr = getZeorAndOne(strs[i-1]);
int zero = arr[0];
int one = arr[1];
for (int j = m; j >= 0; j--) {
for (int k = n; k >= 0; k--) {
if (zero <= j && one <= k){
dp[j][k] = Math.max(dp[j][k], dp[j - zero][k - one] + 1);
}
}
}
}
return dp[m][n];
}

public int[] getZeorAndOne(String str){
int[] arr = new int[2];
int i = 0;
int j = 0;
for (int k = 0; k < str.length(); k++) {
if (str.charAt(k) == '0'){
i++;
}else {
j++;
}
}
arr[0] = i;
arr[1] = j;
return arr;
}
}