前两周一直是一道，都不好意思写了。这周终于题目简单了，刷了三道。
1002. Find Common Characters
Given an arrayAof strings made only from lowercase letters, return a list of all characters that show up in all strings within the list (including duplicates). For example, if a character occurs 3 times in all strings but not 4 times, you need to include that character three times in the final answer.
You may return the answer in any order.
Example 1:
Input: ["bella","label","roller"]
Output: ["e","l","l"] Example 2:
Input: ["cool","lock","cook"]
Output: ["c","o"]
Note:
- 1 <= A.length <= 100
- 1 <= A[i].length <= 100
- A[i][j]is a lowercase letter
题目大意：给你一个字符串数组，让你输出所有字符串都包含的小写字母，都出现了几次就输出几次。
思路：因为只有小写字母，所以可以将所以字符串字母的出现次数统计出来，然后直接遍历输出就好。就是代码有点难写，可能是我还没有领悟到更好的编写方式，浪费了我大量的时间（差不多一个多小时）。
代码：
class Solution { public List<String> commonChars(String[] A) { List<String> list = new ArrayList<>(); List<Map<Character, Integer> > maps = new ArrayList<>(); for(String str : A ) { if( null != str ) { int len = str.length(); Map<Character, Integer> map = new HashMap<>(); for(int i=0; i<len; i++) { char ch = str.charAt(i); Integer cnt = map.get(ch); if( null == cnt ) cnt = Integer.valueOf(0); cnt ++; map.put(ch, cnt); } maps.add(map); } } for(char ch = 'a'; ch<='z'; ch ++) { int last = 105; for(Map<Character, Integer> m : maps ) { Integer cnt = m.get(ch); if( null == cnt ) cnt = Integer.valueOf(0); if( cnt < last ) last = cnt; } while( last != 105 && last > 0 ) { list.add(""+ch); last --; } } return list; } }
1003. Check If Word Is Valid After Substitutions
We are given that the string"abc"is valid.
From any valid stringV, we may splitVinto two piecesXandYsuch thatX + Y(Xconcatenated withY) is equal toV. (XorYmay be empty.) Then,X + "abc" + Yis also valid.
If for exampleS = "abc", then examples of valid strings are:"abc", "aabcbc", "abcabc", "abcabcababcc". Examples of invalid strings are:"abccba","ab","cababc","bac".
Returntrueif and only if the given stringSis valid.
Example 1:
Input: "aabcbc"
Output: true Explanation: We start with the valid string "abc". Then we can insert another "abc" between "a" and "bc", resulting in "a" + "abc" + "bc" which is "aabcbc". Example 2:
Input: "abcabcababcc"
Output: true Explanation: "abcabcabc" is valid after consecutive insertings of "abc". Then we can insert "abc" before the last letter, resulting in "abcabcab" + "abc" + "c" which is "abcabcababcc". Example 3:
Input: "abccba"
Output: false Example 4:
Input: "cababc"
Output: false
Note:
- 1 <= S.length <= 20000
- S[i]is'a','b', or'c'
题目大意：给出一个基础合法串 abc ，然后不断的将 abc 插入到不同的位置，也就形成了更多的合法串。比如：插入到第二个位置变成 aabcbc 这个也是一个合法串。
思路：最开始想到的就是，按照这么变的话，这些串肯定都是固定的，可以先将这些串都求出来然后再对比。后面一想这样可能耗时更长，于是放弃这个思路。然后仔细一想，如果一个串是合法的，那么我如果一直去掉 abc 串，那么后面肯定就剩下空串了。
代码：
class Solution { public boolean isValid(String S) { String valid = "abc"; while( S.lastIndexOf(valid)!=-1 ) { int begin = S.indexOf(valid); String s1 = S.substring(0, begin); String s2 = S.substring(begin+3); S = s1 + s2; if( null == S || "".equals(S) ) { return true; } } return false; } }
1004. Max Consecutive Ones III
Given an arrayAof 0s and 1s, we may change up toKvalues from 0 to 1.
Return the length of the longest (contiguous) subarray that contains only 1s.
Example 1:
Input: A = [1,1,1,0,0,0,1,1,1,1,0], K = 2 Output: 6 Explanation: [1,1,1,0,0,1,1,1,1,1,1] Bolded numbers were flipped from 0 to 1. The longest subarray is underlined. Example 2:
Input: A = [0,0,1,1,0,0,1,1,1,0,1,1,0,0,0,1,1,1,1], K = 3 Output: 10 Explanation: [0,0,1,1,1,1,1,1,1,1,1,1,0,0,0,1,1,1,1] Bolded numbers were flipped from 0 to 1. The longest subarray is underlined.
Note:
- 1 <= A.length <= 20000
- 0 <= K <= A.length
- A[i]is0or1
题目大意：给出一个只包含 0 和 1的数组，然后给出 一个数字K表示你可以将K个0变成1，让你求最长的连续1字串长度。
题目思路：尺取法，不解释。
class Solution { public int longestOnes(int[] A, int K) { int res = 0; int len = A.length; int end = 0, begin = 0; for(int i=0; i<len; i++) { if( A[i] == 0 ) { if( K == 0 ) { if( res < end-begin ) res = end-begin; while( A[begin] == 1 ) begin ++; begin ++; } else K --; } end ++; } return res>end-begin?res:end-begin; } }
1000. Minimum Cost to Merge Stones
There areNpiles of stones arranged in a row. Thei-th pile hasstones[i]stones.
A move consists of merging exactlyKconsecutive piles into one pile, and the cost of this move is equal to the total number of stones in theseKpiles.
Find the minimum cost to merge all piles of stones into one pile. If it is impossible, return-1.
Example 1:
Input: stones = [3,2,4,1], K = 2 Output: 20 Explanation: We start with [3, 2, 4, 1]. We merge [3, 2] for a cost of 5, and we are left with [5, 4, 1]. We merge [4, 1] for a cost of 5, and we are left with [5, 5]. We merge [5, 5] for a cost of 10, and we are left with [10]. The total cost was 20, and this is the minimum possible. Example 2:
Input: stones = [3,2,4,1], K = 3 Output: -1 Explanation: After any merge operation, there are 2 piles left, and we can't merge anymore. So the task is impossible. Example 3:
Input: stones = [3,5,1,2,6], K = 3 Output: 25 Explanation: We start with [3, 5, 1, 2, 6]. We merge [5, 1, 2] for a cost of 8, and we are left with [3, 8, 6]. We merge [3, 8, 6] for a cost of 17, and we are left with [17]. The total cost was 25, and this is the minimum possible.
Note:
- 1 <= stones.length <= 30
- 2 <= K <= 30
- 1 <= stones[i] <= 100
其实就是石子归并，只不过合并的堆数从2变成了k。还是很简单的，可惜第一题费了我太多时间了，不然应该可以做出来。
class Solution { public int mergeStones(int[] a, int K) { int n = a.length; if(n % (K-1) != 1 % (K-1)){ return -1; } int[] cum = new int[n+1]; for(int i = 0;i < n;i++){ cum[i+1] = cum[i] + a[i]; } int[][] dp = new int[n][n]; for(int len = K;len <= n;len++){ for(int i = 0;i+len-1 < n;i++){ int j = i+len-1; int cost = Integer.MAX_VALUE; int ulen = (len-1)/(K-1)*(K-1)+1; if(ulen % (K-1) == 1 %(K-1)){ ulen -= K-1; } int s = len % (K-1) == 1 % (K-1) ? len-(K-1) : len; for(int k = i+1;k <= j;k++){ if((k-i-1)/(K-1) + (j-k+1-1)/(K-1) == (s-1)/(K-1)){ int c = 0; if(i <= k-1)c += dp[i][k-1]; if(k <= j)c += dp[k][j]; cost = Math.min(cost, c); } } if(len % (K-1) == 1 % (K-1)){ cost += cum[j+1] - cum[i]; } dp[i][j] = cost; } } return dp[0][n-1]; } }
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