Comments (4)
/**
* @param {string[]} strs
* @param {number} m
* @param {number} n
* @return {number}
* F(i,C) = max(F(i-1,C),F(i-1,C-w)+v); 0,1背包公式
*
* 多维版
* F(i,m,n) = max(F(i-1,m,n),F(i-1,m-j,n-k)+1); j=当前字符串的0数量,k为1的数量
*
* 每遍历到strs的一个位置时,有两种选择,选择当前字符串,或者不选择
* 如果选择,那么当前F(i,m,n)的最优结果等于减去这个字符串占用的01数量后的最优结果,即F(i-1,m-j,n-k)+1
* 如果不选择,那么当前F(i,m,n)的最优结果等于遍历到上一个字符串时的结果,即F(i-1,m,n)
*
*/
var findMaxForm = function (strs, m, n) {
if (strs.length == 0) return 0;
var dp = Array.from({
length: m + 1
}, () => new Array(n + 1).fill(0));
for (var i = 0; i < strs.length; i++) {
var count1 = 0,
count0 = 0;
for (var item of strs[i]) {
if (item == '1') count1++;
else count0++;
}
for (var k = m; k >= count0; k--) {
for (var j = n; j >= count1; j--) {
dp[k][j] = Math.max(dp[k][j], dp[k - count0][j - count1] + 1);
}
}
}
return dp[m][n];
};
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认领
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我试试english version?
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我试试english version?
好的
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