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  • LeetCode 1049. Last Stone Weight II

    原题链接在这里:https://leetcode.com/problems/last-stone-weight-ii/

    题目:

    We have a collection of rocks, each rock has a positive integer weight.

    Each turn, we choose any two rocks and smash them together.  Suppose the stones have weights x and y with x <= y.  The result of this smash is:

    • If x == y, both stones are totally destroyed;
    • If x != y, the stone of weight x is totally destroyed, and the stone of weight y has new weight y-x.

    At the end, there is at most 1 stone left.  Return the smallest possible weight of this stone (the weight is 0 if there are no stones left.)

    Example 1:

    Input: [2,7,4,1,8,1]
    Output: 1
    Explanation: 
    We can combine 2 and 4 to get 2 so the array converts to [2,7,1,8,1] then,
    we can combine 7 and 8 to get 1 so the array converts to [2,1,1,1] then,
    we can combine 2 and 1 to get 1 so the array converts to [1,1,1] then,
    we can combine 1 and 1 to get 0 so the array converts to [1] then that's the optimal value.

    Note:

    1. 1 <= stones.length <= 30
    2. 1 <= stones[i] <= 100

    题解:

    If we choose any two rocks, we could divide the rocks into 2 groups.

    And calculate the minimum diff between 2 groups' total weight.

    Since stones.length <= 30, stone weight <= 100, maximum total weight could be 3000.

    Let dp[i] denotes whether or not smaller group weight could be i. smaller goupe total weight is limited to 1500. Thus dp size is 1501. It becomes knapsack problem.

    dp[0] = true. We don't need to choose any stone, and we could get group weight as 0.

    For each stone, if we choose it we track dp[i-w] in the previoius iteration. If we don't choose it, track dp[i] from last iteration. Thus it is iterating from big to small.

    Time Complexity: O(n*sum). n = stones.length. sum is total weight of stones.

    Space: O(sum).

    AC Java: 

     1 class Solution {
     2     public int lastStoneWeightII(int[] stones) {
     3         if(stones == null || stones.length == 0){
     4             return 0;
     5         }
     6         
     7         int sum = 0;
     8         boolean [] dp = new boolean[1501];
     9         dp[0] = true;
    10         
    11         for(int w : stones){
    12             sum += w;
    13             
    14             for(int i = Math.min(sum, 1501); i>=w; i--){
    15                 dp[i] = dp[i] | dp[i-w];
    16             }
    17         }
    18         
    19         for(int i = sum/2; i>=0; i--){
    20             if(dp[i]){
    21                 return sum-i-i;
    22             }
    23         }
    24         
    25         return 0;
    26     }
    27 }

    Last Stone Weight进阶.

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  • 原文地址:https://www.cnblogs.com/Dylan-Java-NYC/p/11785116.html
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