[刷题]Word Break

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[LintCode]Word Break

public class Solution {    /**     * @param s: A string s     * @param dict: A dictionary of words dict     */    public boolean wordSegmentation(String s, Set<String> dict) {        // 2015-05-18 假设contains方法时间复杂度为常数，O(n^2)        if (s == null) {            return false;        }        if (s.length() == 0) {            return true;        }        int n = s.length();        int maxLen = getMaxLen(dict); //这样做是为了进一步降低时间复杂度，否则超时        boolean[] can = new boolean[n + 1]; // length is n + 1         can[0] = true;        for (int i = 1; i < n + 1; i++) {            can[i] = false;            for (int len = 1; len <= maxLen && len <= i; len++) {                // 只要有一次不满足条件，直接结束for-loop                if (can[i - len] && dict.contains(s.substring(i - len, i))) {                    can[i] = true;                    break;                }            }// for        }// for        return can[n];    }    // 假定dict有m个词，O(m)    private int getMaxLen(Set<String> dict) {        int maxLen = 0;        for (String word : dict) {            maxLen = Math.max(maxLen, word.length());        }        return maxLen;    }}

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