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LeetCode-Word Break

时间:2015-03-13 23:53:03      阅读:403      评论:0      收藏:0      [点我收藏+]

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Given a string s and a dictionary of words dict, determine if s can be segmented into a space-separated sequence of one or more dictionary words.

For example, given
s = "leetcode",
dict = ["leet", "code"].

Return true because "leetcode" can be segmented as "leet code".

一看这题目,就感觉里面回溯法,因为这种题,在回溯法中在平常不过了。上代码:

public boolean wordBreak(String s, Set<String> dict) {
    	StringBuilder sb = new StringBuilder();
        return helper(s, dict, 0, sb);
    }
    private boolean helper(String s, Set<String> dict, int index, StringBuilder sb) {
    	if (index == dict.size()) {
    		return s.equals(sb.toString());
    	}
    	Iterator<String> iter = dict.iterator();
    	while (iter.hasNext()) {
    		StringBuilder sb1 = new StringBuilder(sb);
    		String dic = iter.next();
    		sb.append(dic);
    		iter.remove();
    		if (helper(s, dict, index+1, sb))
    			return true;
    		dict.add(dic);
    		sb = sb1;
    	}
    	return false;
    }

结果超时!一般情况下,对于回溯的时间限制都是比较宽的,但是超时!肯定有优化的空间吧,剪枝啊:、

public boolean wordBreak(String s, Set<String> dict) {
    	StringBuilder sb = new StringBuilder();
        return helper(s, dict, 0, sb);
    }
    private boolean helper(String s, Set<String> dict, int index, StringBuilder sb) {
    	if (!s.startsWith(sb.toString())) {
    		return false;
    	}
    	if (index == dict.size()) {
    		return s.equals(sb.toString());
    	}
    	Iterator<String> iter = dict.iterator();
    	while (iter.hasNext()) {
    		StringBuilder sb1 = new StringBuilder(sb);
    		String dic = iter.next();
    		sb.append(dic);
    		iter.remove();
    		if (helper(s, dict, index+1, sb))
    			return true;
    		dict.add(dic);
    		sb = sb1;
    	}
    	return false;
    }
其实这个剪枝可以减去很多分支,但是还是超时!怎么办?只能用最高效的动态规划了,上代码:

public boolean wordBreak1(String s, Set<String> dict) {
    	int length = s.length();
        boolean[] can = new boolean[length+1];
        can[0] = true;
        for (int i = 1; i <= length; i++) {
            for (int j = 0; j < i; j++) {
                if (can[j] && dict.contains(s.substring(j, i))) {
                    can[i] = true;
                    break;
                }
            }
        }
        return can[length];
    }
boolean数组can[i]是指,s.substring(0,i)是可以分割的,所示数组最后一个元素即为所求。






LeetCode-Word Break

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原文地址:http://blog.csdn.net/my_jobs/article/details/44245765

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