473. Matchsticks to Square

Remember the story of Little Match Girl? By now, you know exactly what matchsticks the little match girl has, please find out a way you can make one square by using up all those matchsticks. You should not break any stick, but you can link them up, and each matchstick must be used exactly one time.

Your input will be several matchsticks the girl has, represented with their stick length. Your output will either be true or false, to represent whether you could make one square using all the matchsticks the little match girl has.

Example 1:

Input: [1,1,2,2,2]Output: trueExplanation: You can form a square with length 2, one side of the square came two sticks with length 1.

Example 2:

Input: [3,3,3,3,4]Output: falseExplanation: You cannot find a way to form a square with all the matchsticks.

Note:

1. The length sum of the given matchsticks is in the range of 0 to 10^9.
2. The length of the given matchstick array will not exceed 15.

思路：最开始想用BFS，遍历完最后看是否出现过sum/4,2*sum/4,3*sum/4

wrong idea : 考虑1,1,4,4,2,3,2,3如果2,2,1组合就没戏唱了

package l473;import java.util.HashSet;import java.util.Set;/* * 想用完全背包 * 仔细一想其实并不需要那么复杂 *  * 分成2步 * 1. 把数组分成和相等的2部分 * 2. 把2部分再分别等分 * 如果2步都OK，那就return true *  * 其实一步就好，BFS，每个数据都可取或者不取 * 最后看是否出现过sum/4,2*sum/4,3*sum/4 *  * wrong idea : 考虑1,1,4,4,2,3,2,3如果2,2,1组合就没戏唱了 *  * The length of the given matchstick array will not exceed 15.或者还可以穷举? */public class BFS_WA {    public boolean makesquare(int[] nums) {    if(nums == null || nums.length == 0)return false;    int sum = 0;    for(int i : nums)sum+=i;    if(sum % 4 != 0)return false;        Set<Integer> s = new HashSet<Integer>();    Set<Integer> ss = new HashSet<Integer>();    s.add(0);        // bfs    for(int i : nums) {    if(i > sum/4)return false;    for(int j : s)    ss.add(j+i);    s.addAll(ss);    ss.clear();    }            return s.contains(sum/4) && s.contains(sum/2) && s.contains(sum/4*3);    }}

那就DFS，先降序排列会提高速度，因为数据大先填的坑就大，后面只能填一些小的，需要遍历的情况就更小

 

package l473;import java.util.Arrays;/* * DFS * 先降序排列会提高速度，因为数据大先填的坑就大，后面只能填一些小的，需要遍历的情况就更小 */public class Solution {int[] edge =  new int[4];boolean f = false;    public boolean makesquare(int[] nums) {    if(nums == null || nums.length == 0)return false;    int sum = 0;    for(int i : nums)sum+=i;    if(sum % 4 != 0)return false;        Arrays.sort(nums);    dfs(nums, sum/4, nums.length-1);            return f;    }private void dfs(int[] nums, int max, int s) {if(f)return;if(s == -1) {if(edge[0]==max && edge[1]==max && edge[2]==max) {f = true;return;}}// a little more pruning to AC if not sortfor(int i=0; i<4 && !f; i++) {if(edge[i]+nums[s] <= max) {edge[i] += nums[s];dfs(nums, max, s-1);edge[i] -= nums[s];}}}}