Leetcode - Last Stone Weight II

https://leetcode.com/problems/last-stone-weight-ii/

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.

func lastStoneWeightII(stones []int) int {
    sum := 0
    for _, s := range stones {
        sum += s
    }
    n := len(stones)
    dp := make([][]bool, sum+1)
    for i := 0; i < len(dp); i++ {
        dp[i] = make([]bool, n+1)
    }
    for i := 0; i <= n; i++ {
        dp[0][i] = true
    }
    s := 0
    for i := 1; i <= n; i++ {
        for j := 1; j <= sum/2; j++ {
            if dp[j][i-1] || (j >= stones[i-1] && dp[j - stones[i-1]][i-1]) {
                dp[j][i] = true
                if j > s {
                    s = j
                }
            }
        }
    }
    return sum - s*2
}

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