leetcode 子数组范围和

给你一个整数数组 nums 。nums 中,子数组的 范围 是子数组中最大元素和最小元素的差值。

返回 nums 中 所有 子数组范围的  

子数组是数组中一个连续 非空 的元素序列。

示例 1:

输入:nums = [1,2,3]
输出:4
解释:nums 的 6 个子数组如下所示:
[1],范围 = 最大 - 最小 = 1 - 1 = 0 
[2],范围 = 2 - 2 = 0
[3],范围 = 3 - 3 = 0
[1,2],范围 = 2 - 1 = 1
[2,3],范围 = 3 - 2 = 1
[1,2,3],范围 = 3 - 1 = 2
所有范围的和是 0 + 0 + 0 + 1 + 1 + 2 = 4

示例 2:

输入:nums = [1,3,3]
输出:4
解释:nums 的 6 个子数组如下所示:
[1],范围 = 最大 - 最小 = 1 - 1 = 0
[3],范围 = 3 - 3 = 0
[3],范围 = 3 - 3 = 0
[1,3],范围 = 3 - 1 = 2
[3,3],范围 = 3 - 3 = 0
[1,3,3],范围 = 3 - 1 = 2
所有范围的和是 0 + 0 + 0 + 2 + 0 + 2 = 4

示例 3:

输入:nums = [4,-2,-3,4,1]
输出:59
解释:nums 中所有子数组范围的和是 59

提示:

  • 1 <= nums.length <= 1000
  • -109 <= nums[i] <= 109

java

class Solution {
    public long subArrayRanges(int[] nums) {
        int n = nums.length;
        long res = 0;
        for (int i = 0; i < n; i++) {
            int min_val = nums[i];
            int max_val = nums[i];
            for (int j = i; j < n; j++) {
                min_val = Math.min(min_val, nums[j]);
                max_val = Math.max(max_val, nums[j]);
                res += max_val - min_val;
            }
        }
        return res;
    }
}

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Origin blog.csdn.net/qq_27060423/article/details/121889999