題目

A string of '0’s and '1’s is monotone increasing if it consists of some number of '0’s (possibly 0), followed by some number of '1’s (also possibly 0.)

We are given a string S of '0’s and '1’s, and we may flip any ‘0’ to a ‘1’ or a ‘1’ to a ‘0’.

Return the minimum number of flips to make S monotone increasing.

Example 1:

Input: "00110"
Output: 1
Explanation: We flip the last digit to get 00111.

Example 2:

Input: "010110"
Output: 2
Explanation: We flip to get 011111, or alternatively 000111.

Example 3:

Input: "00011000"
Output: 2
Explanation: We flip to get 00000000.

Note:

1. 1 <= S.length <= 20000
2. S only consists of ‘0’ and ‘1’ characters.

題目大意

給定字串 由 0,1組成，求 換最少的 字元使得 字串 不遞減。

思路

方法一 動態規劃

狀態： dp[i][0] str[0:i] 最後字元為 0 時，最少 換字元數。 dp[i][1] str[0:i] 最後字元為 1 時，最少 換字元數。

狀態轉移方程：

            if(S.charAt(i-1) == '0'){
                dp[i][0] = dp[i-1][0];
                dp[i][1] = dp[i-1][1]+1;
            }
            else{
                dp[
 
i][0] = dp[i-1][0]+1;
                dp[i][1] = Math.min(dp[i-1][1],dp[i-1][0]);
            }

code

class Solution {
    public int minFlipsMonoIncr(String S) {
        int strlen = S.length();
        int[][] dp = new int[strlen+1][2];
        dp[0][0] = dp[0][1] = 0;
        for(int i = 1;i<=strlen;i++){
            if(S.charAt(i-1) == '0'){
                dp[i][0] = dp[i-1][0];
                dp[i][1] = dp[i-1][1]+1;
            }
            else{
                dp[i][0] = dp[i-1][0]+1;
                dp[i][1] = Math.min(dp[i-1][1],dp[i-1][0]);
            }
        }
        return Math.min(dp[strlen][0],dp[strlen][1]);
    }
}

由於dp[i]只用到 前一個 dp[i-1]，可以 簡化dp，使用一維dp。

class Solution {
    public int minFlipsMonoIncr(String S) { 
        int[] dp = new int[2];
        for(int i = 0;i<S.length();i++){
            if(S.charAt(i)=='0')
                dp[1] = dp[1]+1; 
            else{
                dp[1] = Math.min(dp[1],dp[0]);
                dp[0] = dp[0] + 1;
            }
        }
        return Math.min(dp[0],dp[1]);
    }
}