給一個(gè)數(shù)組内颗，要求你盡可能多的切割這個(gè)數(shù)組袖迎，使得每一個(gè)小段分別sort之后姻报，整個(gè)數(shù)組就sort了
if minValue of Latter chunk is larger than the max value of former chunk, there can be a slice.
traverse the array from end to beginning
class Solution {
public int getChunks (int[] nums) {
List<Integer> list = new ArrayList<>();
if (nums == null || nums.length == 0) {
return -1;
}
int min = nums[nums.length - 1];
for (int i = nums.length - 1; i >= 0; i--) {
if (nums[i] > min) {
continue;
}
list.add(i + 1);
min = nums[i];
} // end of for
return list.size() + 1;
}
}
