剑指 Offer 51. 数组中的逆序对

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class Solution {
    public int reversePairs(int[] record) {
        int[] tmp = new int[record.length];
        return mergeSort(record, tmp,0, record.length - 1);
    }

    private int mergeSort(int[] record, int[] tmp, int left, int right) {
        if (left >= right) {
            // 不要遗漏此处终止条件
            return 0;
        }

        int mid = (left + right) >>> 1;
        // split to: [left, mid] / [mid + 1, right]
        int reversePairs = 0;
        reversePairs += mergeSort(record, tmp, left, mid);
        reversePairs += mergeSort(record, tmp, mid + 1, right);

        for (int i = left; i <= right; i++) {
            tmp[i] = record[i];
        }

        int i = left, j = mid + 1;
        for (int k = left; k <= right; k++) {
            if (i == mid + 1) {
                record[k] = tmp[j++];
            } else if (j == right + 1) { // 注意此处是与 right 进行比较而不是 record.length
                record[k] = tmp[i++];
            } else if (tmp[i] <= tmp[j]) {
                record[k] = tmp[i++];
            } else {
                // [1, 3, 5] / [1, 1, 2]
                //     i           j
                record[k] = tmp[j++];
                reversePairs += mid - i + 1; // 累加左侧大于 tmp[j] 的元素个数
            }
        }

        return reversePairs;
    }
}

References

剑指 Offer 51. 数组中的逆序对