红和绿

#include<stdio.h>
#include<string.h>
#define min(a,b) a<b?a:b
int main(){
    char s[100];
    scanf("%s",s);
    int i,j,Min=1e8,n=strlen(s);
    for(i=0;i<n;i++){
        int cnt=0;
        for(j=0;j<i;j++) if(s[j]!='R') cnt++;
        for(j=i;j<n;j++) if(s[j]!='G') cnt++;
        Min=min(Min,cnt);
    }
    printf("%d\n",Min);
}

import java.util.*;
public class Main {

    public static void main(String[] args) {
        // TODO Auto-generated method stub
        Scanner sc=new Scanner(System.in);
        String s=sc.nextLine();
        int ans=Integer.MAX_VALUE;
        for(int i=0;i<s.length();i++)
        {
            int r=0;
            for(int j=0;j<i;j++)
            {
                if(s.charAt(j)!='R')
                    r++;
            }
            for(int k=i;k<s.length();k++)
            {
                if(s.charAt(k)!='G')
                    r++;
                
            }
            ans=Math.min(ans, r);
        }
        System.out.println(ans);
    }

}

	

import java.util.Scanner;
public class Main {
    public static void main(String[] args) {
        Scanner sc = new Scanner(System.in);
        String s = sc.nextLine();
        System.out.println(getColorWayPreprocess(s));
    }
    public static int getColorWayPreprocess(String s) {
        char[] chs = s.toCharArray();
        int n = chs.length;
        int[] leftG = new int[n];
        int[] rightR = new int[n];
        for (int i = 0; i < n; i++) {
            if (chs[i] == 'G') {
                leftG[i] = i == 0 ? 1 : leftG[i - 1] + 1;
            } else {
                leftG[i] = i == 0 ? 0 : leftG[i - 1];
            }
        }
        for (int i = n - 1; i >= 0; i--) {
            if (chs[i] == 'R') {
                rightR[i] = i == n - 1 ? 1 : rightR[i + 1] + 1;
            } else {
                rightR[i] = i == n - 1 ? 0 : rightR[i + 1];
            }
        }
        int res = Integer.MAX_VALUE;
        for (int r = -1; r < n; r++) {
            int tmp;
            if (r == -1) {
                tmp = rightR[0];
            } else if (r == n - 1) {
                tmp = leftG[n - 1];
            } else {
                tmp = leftG[r] + rightR[r + 1];
            }
            res = Math.min(res, tmp);
        }
        return res;
    }
}

使用预先构建数组, 使得时间复杂度为O(n)