561. Array Partition I(Java)

本文介绍了一种算法,旨在将给定的2n整数数组通过最优配对方式,使得每对中较小数的总和最大化。示例展示了如何通过排序和选择偶数位置元素来实现这一目标。

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给定一个2n整数数组,您的任务是将这些整数分组为n对整数,例如(a1,b1),(a2,b2),…,(an,bn),它使所有i的min(ai,bi)之和尽可能大,从1到n。              

例1:              输入:[1,4,3,2]              产量:4              说明:n为2,对的最大和为4=min(1,2)+min(3,4)。            

class Solution {
    public int arrayPairSum(int[] nums) {
        int res = 0;
        Arrays.sort(nums);
        for(int i=0; i<nums.length;i+=2){
            res =res+nums[i];
        }
        return res;
    }
}

 注:              n是一个正整数,在[10000]的范围内。              数组中的所有整数都将在范围[-10000,10000]内。 

``` package oy.tol.tra; import java.util.function.Predicate; public class Algorithms { public static <T extends Comparable<T>> void reverse(T[] array) { int i = 0; int j = array.length - 1; while (i < j) { T temp = array[i]; array[i] = array[j]; array[j] = temp; i++; j--; } } public static <T extends Comparable<T>> void sort(T[] array) { boolean swapped; for (int i = 0; i < array.length - 1; i++) { swapped = false; for (int j = 0; j < array.length - 1 - i; j++) { if (array[j].compareTo(array[j + 1]) > 0) { T temp = array[j]; array[j] = array[j + 1]; array[j + 1] = temp; swapped = true; } } if (!swapped) break; } } public static <T extends Comparable<T>> int binarySearch(T aValue, T[] fromArray, int fromIndex, int toIndex) { int low = fromIndex; int high = toIndex ; while (low <= high) { int mid = low + (high - low) / 2; int cmp = fromArray[mid].compareTo(aValue); if (cmp < 0) { low = mid + 1; } else if (cmp > 0) { high = mid - 1; } else { return mid; } } return -1; } public static <E extends Comparable<E>> void fastSort(E[] array) { quickSort(array, 0, array.length - 1); } private static <E extends Comparable<E>> void quickSort(E[] array, int begin, int end) { if (begin < end) { int partitionIndex = partition(array, begin, end); quickSort(array, begin, partitionIndex - 1); quickSort(array, partitionIndex + 1, end); } } public static <T> int partition(T[] array, int count, Predicate<T> rule) { if (array == null || count <= 0) { return 0; // No elements to partition } int left = 0; // Pointer for elements that satisfy the rule int right = count - 1; // Pointer for elements that do not satisfy the rule while (left <= right) { // Move the left pointer to the right until an element does not satisfy the rule while (left <= right && rule.test(array[left])) { left++; } // Move the right pointer to the left until an element satisfies the rule while (left <= right && !rule.test(array[right])) { right--; } // Swap elements if pointers have not crossed if (left <= right) { T temp = array[left]; array[left] = array[right]; array[right] = temp; left++; right--; } } // Return the index of the first element that does not satisfy the rule return left; } }```The method partition(T[], int, Predicate<T>) in the type Algorithms is not applicable for the arguments (E[], int, int)Java(67108979)
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03-11
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