区间和的个数
题目
给你一个整数数组 nums 以及两个整数 lower 和 upper 。求数组中,值位于范围 [lower, upper] (包含 lower 和 upper)之内的 区间和的个数 。
区间和 S(i, j) 表示在 nums 中,位置从 i 到 j 的元素之和,包含 i 和 j (i ≤ j)。
示例
输入:nums = [-2,5,-1], lower = -2, upper = 2
输出:3
解释:存在三个区间:[0,0]、[2,2] 和 [0,2] ,对应的区间和分别是:-2 、-1 、2 。
题解
动态规划
挂了,超时~~~
var countRangeSum = function(nums, lower, upper) {
const len = nums.length;
let result = 0;
const dp = Array(len)
for (var i = 0; i < len; i++){
dp[i] = nums[i];
if (dp[i] >= lower && dp[i] <= upper) result++
}
for (var i = 0; i < len-1; i++){
for (var j = i+1; j < len; j++) {
dp[i] += dp[j];
if (dp[i] >= lower && dp[i] <= upper) result++
}
}
return result
};
代码
const countRangeSumRecursive = (sum, lower, upper, left, right) => {
if (left === right) {
return 0;
} else {
const mid = Math.floor((left + right) / 2);
const n1 = countRangeSumRecursive(sum, lower, upper, left, mid);
const n2 = countRangeSumRecursive(sum, lower, upper, mid + 1, right);
let ret = n1 + n2;
// 首先统计下标对的数量
let i = left;
let l = mid + 1;
let r = mid + 1;
while (i <= mid) {
while (l <= right && sum[l] - sum[i] < lower) l++;
while (r <= right && sum[r] - sum[i] <= upper) r++;
ret += (r - l);
i++;
}
// 随后合并两个排序数组
const sorted = new Array(right - left + 1);
let p1 = left, p2 = mid + 1;
let p = 0;
while (p1 <= mid || p2 <= right) {
if (p1 > mid) {
sorted[p++] = sum[p2++];
} else if (p2 > right) {
sorted[p++] = sum[p1++];
} else {
if (sum[p1] < sum[p2]) {
sorted[p++] = sum[p1++];
} else {
sorted[p++] = sum[p2++];
}
}
}
for (let i = 0; i < sorted.length; i++) {
sum[left + i] = sorted[i];
}
return ret;
}
}
var countRangeSum = function(nums, lower, upper) {
let s = 0;
const sum = [0];
for(const v of nums) {
s += v;
sum.push(s);
}
return countRangeSumRecursive(sum, lower, upper, 0, sum.length - 1);
};