1531. 压缩字符串 II（动态规划）

class Solution {
    public int getLengthOfOptimalCompression(String s, int k) {
        int n = s.length();
        int[][] dp = new int[n+1][k+1]; // dp[i][j]:考虑前i个字符最多删除j个的最小长度
        for(int[] d : dp) Arrays.fill(d,Integer.MAX_VALUE >> 1);
        dp[0][0] = 0;
        for(int i = 1; i <= n; i++) {
            for(int j = 0; j <= k && j <= i; j++) {
                if(j > 0) dp[i][j] = dp[i-1][j-1];  // 情况1:删除第i个字符
                // 情况2：保留第i个字符
                int same = 0, diff = 0;
                for(int f = i; f >= 1 && diff <= j; f--) { // 向前遍历与第i个字符相同的字符则考虑保留
                    if(s.charAt(f-1) == s.charAt(i-1)) {
                        dp[i][j] = Math.min(dp[i][j],dp[f-1][j-diff] + calc(++same));
                    } else {
                        diff++;
                    }
                }
            }
        }
        return dp[n][k];
    }

    public int calc(int n) {
        if(n == 1) return 1;
        if(n < 10) return 2;
        if(n < 100) return 3;
        return 4;
    }
}