题目

思路
O(n)的思路比较简单，直接用两个下标扫一遍即可；
O(nLogn)有点难，个人感觉应该是先得到Sum[i]（前i+1）个数的和，因为数字都是正数，那么Sum数组可以用二分查找。我们扫一遍Sum，再二分查找符合条件的前一个Sum的位置即可。<??http://www.2cto.com/kf/ware/vc/" target="_blank" class="keylink">vcD4NCjxwPjxzdHJvbmc+tPrC6zwvc3Ryb25nPjxiciAvPg0KTyhuKaO6PC9wPg0KPHByZSBjbGFzcz0="brush:java;"> int minSubArrayLen(int s, int * nums, int numsSize) { int sum = nums[0], head = 0, tail = 0, minL = numsSize + 1; while (tail < numsSize) { if (tail - head + 1 < minL && sum >= s) minL = tail - head + 1; if (sum >= s) sum -= nums[head++]; else sum += nums[++tail]; if (head > tail) tail = head; } return minL == numsSize + 1 ? 0 : minL; }

O(nLogn)：

// 在不下降的序列中寻找恰好比target小的数出现位置，也即最后一个比target小的数出现的位置
// search a number that is exactly less than 'target', which means the last number less than 'target'
int binarySearchIncreaseLastSmaller(int l, int r, int target, int * nums) {  
    if (l >= r) return -1;
    while (l < r - 1) {
        int m = l + ((r - l) >> 1);
        if (nums[m] < target) l = m;
        else r = m - 1;
    }
    if (nums[r] < target) return r;
    else if (nums[l] < target) return l;
    else return -1;
}

int minSubArrayLen(int s, int * nums, int numsSize) {
    int * Sum = (int*)malloc(sizeof(int) * (numsSize + 1)), minL = numsSize + 1;
    Sum[0] = 0;
    for (int i = 1; i <= numsSize; i++) Sum[i] = Sum[i - 1] + nums[i - 1];
    for (int i = 1; i <= numsSize; i++) {
        if (Sum[i] >= s) {
            int k = Sum[i];
            int BeforePos = binarySearchIncreaseLastSmaller(0, i, Sum[i] - s + 1, Sum);
            if (BeforePos != -1 && i - BeforePos < minL) minL = i - BeforePos;
        }
    }
    return minL == numsSize + 1 ? 0 : minL;
}