NC_35_EDIT_COST NC_36_findMedianinTwoSortedAray NC_37_MERGEINTERVAL

package org.example.interview.practice;

/**
 * @author xianzhe.ma
 * @date 2021/11/6
 */

public class NC_36_findMedianinTwoSortedAray {

    public int findMedianinTwoSortedAray (int[] arr1, int[] arr2) {
        // write code here
        int finalPos = arr1.length;
        int firstIndex = 0;
        int secondIndex = 0;

        if (finalPos == 1) {
            return Math.min(arr1[0], arr2[0]);
        }
        int tag = -1;//0 first 1 second
        while (firstIndex + secondIndex  < finalPos) {
            if (arr1[firstIndex] <= arr2[secondIndex]) {
                firstIndex++;
                tag = 0;
            } else {
                secondIndex++;
                tag = 1;
            }
        }

        if (tag == 0) {
            return arr1[firstIndex-1];
        } else {
            return arr2[secondIndex-1];
        }
    }
}

package org.example.interview.practice;

import java.util.ArrayList;
import java.util.Comparator;

/**
 * @author xianzhe.ma
 * @date 2021/7/27
 */

public class NC_37_MERGEINTERVAL {

    public static class Interval {
        int start;
        int end;

        Interval() {
            start = 0;
            end = 0;
        }

        Interval(int s, int e) {
            start = s;
            end = e;
        }
    }

    public ArrayList<Interval> merge(ArrayList<Interval> intervals) {
        intervals.sort(new Comparator<Interval>() {
            @Override
            public int compare(Interval o1, Interval o2) {
                return o1.start - o2.start;
            }
        });

        ArrayList<Interval> ans = new ArrayList<Interval>();
        int len = intervals.size();
        if (len == 0) return ans;
        for (int i = 1; i < len; i++) {
            if (intervals.get(i).start <= intervals.get(i - 1).end) {
                intervals.get(i).start = Math.min(intervals.get(i - 1).start, intervals.get(i).start);
                intervals.get(i).end = Math.max(intervals.get(i - 1).end, intervals.get(i).end);
            } else {
                ans.add(intervals.get(i - 1));
            }
        }
        ans.add(intervals.get(len - 1));
        return ans;
    }
}

描述
给定两个字符串str1和str2，再给定三个整数ic，dc和rc，分别代表插入、删除和替换一个字符的代价，请输出将str1编辑成str2的最小代价。

public class NC_35_EDIT_COST{
    /**
     * min edit cost
     * @param str1 string字符串 the string
     * @param str2 string字符串 the string
     * @param ic int整型 insert cost
     * @param dc int整型 delete cost
     * @param rc int整型 replace cost
     * @return int整型
     */
    public int minEditCost (String str1, String str2, int ic, int dc, int rc) {
        // write code here
        int len1 = str1.length();
        int len2 = str2.length();
        //dp[i][j]不包含第i个字符和第j个字符，因此需要+1
        int[][] dp = new int[len1+1][len2+1];
        //矩阵第一行表示空字符串转为str2的代价，插入j个字符
        for(int j = 1; j <= len2; j++) {
            dp[0][j] = j * ic;
        }
        //矩阵第1列表示str1转为空字符串的代价，删除i个字符
        for(int i = 1; i <= len1; i++) {
            dp[i][0] = i * dc;
        }
        for(int i = 1; i <= len1; i++) {
            for(int j = 1; j <= len2; j++) {
                if(str1.charAt(i-1) == str2.charAt(j-1)) {
                    dp[i][j] = dp[i-1][j-1];
                } else {
                    dp[i][j] = Math.min(dp[i][j-1] + ic, dp[i-1][j] + dc);
                    dp[i][j] = Math.min(dp[i][j], dp[i-1][j-1] + rc);
                }
            }
        }
        return dp[len1][len2];
    }
}