动态规划算法，最小代价

最新推荐文章于 2022-07-15 17:35:11 发布

原创最新推荐文章于 2022-07-15 17:35:11 发布 · 2.9k 阅读

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该博客探讨了一个字符串转换问题，通过动态规划算法计算将字符串"AABB"转换为"BB"的最小代价，其中删除操作代价为2，插入操作代价为插入字符数+2。文章展示了具体的Java代码实现。

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<span style="color: rgb(51, 51, 51); font-family: Helvetica, Tahoma, Arial, sans-serif; font-size: 14px; line-height: 24px; background-color: rgb(245, 245, 245);">将一个字符串通过插入字符串或者删除字符串的方式转换为另一个给定的字符串。删除连续n个字符的操作的代价为2，插入n个字符的代价为n+2。求转换的最小代价。</span>

<span style="color: rgb(51, 51, 51); font-family: Helvetica, Tahoma, Arial, sans-serif; font-size: 14px; line-height: 24px; background-color: rgb(245, 245, 245);">
</span>

package com.test;

import com.copy.test.Demo1CopyTest;

public class Demo {

	public static void main(String[] args) {
		GetMinExpenses("AABB", "BB");
	}

	public static int DEL = -1;

	public static int ORIGAL = 0;

	public static int ADD = 1;

	public static int getAddCount(int f, int type) {
		int minCost;
		if (type == ADD) {
			minCost = f + 1;
		} else {
			minCost = f + 3;
		}
		return minCost;
	}

	public static int getDelCount(int f, int type) {
		int minCost = 0;
		if (type == DEL) {
			minCost = f + 0;
		} else {
			minCost = f + 2;
		}
		return minCost;
	}

	public static int getMin(int a, int b) {
		return a < b ? a : b;
	}

	public static int GetMinExpenses(String aString, String bString) {
		int[][] f = new int[aString.length() + 1][bString.length() + 1];//f[i][j] 从a[i] -> b[j]的最小代价
		f[0][0] = 0;
		int operator[][] = new int[aString.length() + 1][bString.length() + 1];//用于记录操作

		for (int i = 1; i < aString.length() + 1; i++) {
			f[i][0] = 2;
			operator[i][0] = DEL;
		}
		for (int i = 1; i < bString.length() + 1; i++) {
			f[0][i] = 2 + i;
			operator[0][i] = ADD;
		}
		int type = ORIGAL;
		for (int i = 1; i < aString.length() + 1; i++) {
			// System.out.println();
			for (int j = 1; j < bString.length() + 1; j++) {
				// System.out.print(minCost + "_" + operator[i][j] + " ");
				int tempType;
                int cost = 0;
                if (aString.charAt(i - 1) != bString.charAt(j - 1))
                {
                    cost = 5;
                }
                int minCost;
                int delCount = getDelCount(f[i - 1][j], operator[i - 1][j]);
                int addCount = getAddCount(f[i][j - 1], operator[i][j - 1]);
                if (delCount >= addCount)
                {
                    operator[i][j] = ADD;
                    minCost = addCount;
                }
                else
                {
                    operator[i][j] = DEL;
                    minCost = delCount;
                }
                if (minCost > f[i - 1][j - 1] + cost)
                {
                    operator[i][j] = ORIGAL;
                    minCost = f[i - 1][j - 1] + cost;
                }
                f[i][j] = minCost;
			}
			// System.out.println("");
		}
		return f[aString.length()][bString.length()];
	}

}