leetcode-1300-转变数组后最接近目标值的数组和

题目描述：

 解法一：二分法  提交：O(nlogn）

class Solution:
    def findBestValue(self, arr: List[int], target: int) -> int:
        def helper(num):
            res = 0
            for i in arr:
                if i > num:
                    res += num
                else:
                    res += i
            return res
        
        l,r = 0,min(target,max(i for i in arr))
        while l + 1 < r:
            mid = l - (l-r)//2
            if helper(mid) < target:
                l = mid
            else:
                r = mid
        if abs(helper(l) - target) > abs(helper(r) - target):
            return r
        else:
            return l

方法二：排序 提交： O(nlogn）

class Solution:
    def findBestValue(self, arr: List[int], target: int) -> int:
        arr.sort()
        n = len(arr)
        tmp = 0
        for i in range(n):
            if tmp + arr[i] * n < target:
                tmp += arr[i]
                n -= 1
            elif tmp + arr[i] * n < target:
                return arr[i]
            else:
                t = target - tmp
                ans = t // n
                if t - ans * n <= (ans + 1) *n - t:
                    return ans
                else:
                    return ans + 1
        return arr[-1]

java：

class Solution {
    public int findBestValue(int[] arr, int target) {
        Arrays.sort(arr);
        int len = arr.length;
        int curSum = 0;
        for (int i = 0; i < len; i++) {
            int curAve = (target - curSum) / (len - i);
            if (curAve <= arr[i]) {
                //即判断curAve两边
                double curAveDou = (target * 1.0 - curSum) / (len - i);
                if (curAveDou - curAve <= 0.5) {
                    return curAve;
                } else {
                    return curAve + 1;
                }
            }
            curSum += arr[i];
        }
        return arr[len - 1];
    }
}

标签：tmp,arr,1300,target,helper,int,res,数组,leetcode
来源： https://www.cnblogs.com/oldby/p/13128540.html