lintcode(32)最小子串覆盖

来源：互联网发布：佳能单反如何测光知乎编辑：程序博客网时间：2024/06/07 00:39

描述：

给定一个字符串source和一个目标字符串target，在字符串source中找到包括所有目标字符串字母的子串。

样例：

给出source = "ADOBECODEBANC"，target = "ABC" 满足要求的解 "BANC"

思路：

先确定是不是子串，然后缩小范围

public class Solution {    /**     * @param source: A string     * @param target: A string     * @return: A string denote the minimum window     *          Return "" if there is no such a string     */    public String minWindow(String source, String target) {        // write your code        int s = source.length();        int t = target.length();        String result = "";        HashMap<Character , Integer> tar = new HashMap<Character , Integer>();        for(int i = 0;i<t;i++){            char ta = target.charAt(i);            if(tar.containsKey(ta)){                tar.put(ta , tar.get(ta) + 1);            }else{                tar.put(ta , 1);            }        }                HashMap<Character , Integer> record = new HashMap<Character , Integer>();        int count = 0;        int right = 0;        boolean change = false;        for(int i = 0;i<s;i++){            char so = source.charAt(i);            if(record.containsKey(so)){                record.put(so , record.get(so) + 1);                if(tar.containsKey(so) && tar.get(so) >= record.get(so)){                    count++;                }            }else{                record.put(so , 1);                if(tar.containsKey(so) && tar.get(so) >= record.get(so)){                    count++;                }            }            if(count == t){                right = i;                change = true;                break;            }        }        if(change){            result = source.substring(0 , right+1);            int left = 0;            char m = result.charAt(left);            while(!tar.containsKey(m) || record.get(m) > tar.get(m)){                if(tar.containsKey(m) && record.get(m) > tar.get(m)){                    record.put(m , record.get(m) - 1);                }                left++;                m = result.charAt(left);            }            result = result.substring(left);        }        return result;    }}

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